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Commit 3e45eec

Browse files
author
liningrui
committed
add new solutions
1 parent cc2e114 commit 3e45eec

20 files changed

+1100
-0
lines changed

‎script/[222]完全二叉树的节点个数.py‎

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Original file line numberDiff line numberDiff line change
@@ -51,4 +51,21 @@
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# self.right = right
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class Solution:
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def countNodes(self, root: TreeNode) -> int:
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def dfs(root):
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if not root:
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return 0
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lh = 0
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rh = 0
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left = root
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right = root
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while left:
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left = left.left
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lh +=1
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while right:
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right = right.right
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rh+=1
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if lh==rh:
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return 2 ** lh - 1
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return dfs(root.left) + dfs(root.right) +1
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return dfs(root)
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[235]二叉搜索树的最近公共祖先.py‎

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@@ -43,5 +43,13 @@
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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l = min(p.val, q.val)
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r = max(p.val, q.val)
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while root and (root.val<l or root.val>r):
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if root.val<l:
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root = root.right
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else:
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root = root.left
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return root
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# leetcode submit region end(Prohibit modification and deletion)
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# 序列化二叉树的一种方法是使用 前序遍历 。当我们遇到一个非空节点时,我们可以记录下这个节点的值。如果它是一个空节点,我们可以使用一个标记值记录,例如 #。
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#
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#
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#
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#
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# 例如,上面的二叉树可以被序列化为字符串 "9,3,4,#,#,1,#,#,2,#,6,#,#",其中 # 代表一个空节点。
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#
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# 给定一串以逗号分隔的序列,验证它是否是正确的二叉树的前序序列化。编写一个在不重构树的条件下的可行算法。
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#
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# 保证 每个以逗号分隔的字符或为一个整数或为一个表示 null 指针的 '#' 。
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#
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# 你可以认为输入格式总是有效的
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#
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#
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# 例如它永远不会包含两个连续的逗号,比如 "1,,3" 。
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#
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#
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# 注意:不允许重建树。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
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# 输出: true
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#
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# 示例 2:
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#
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#
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# 输入: preorder = "1,#"
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# 输出: false
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#
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#
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# 示例 3:
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#
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#
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# 输入: preorder = "9,#,#,1"
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# 输出: false
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#
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= preorder.length <= 104
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# preorder 由以逗号 "," 分隔的 [0,100] 范围内的整数和 "#" 组成
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#
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# Related Topics 栈 树 字符串 二叉树 👍 393 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def isValidSerialization(self, preorder: str) -> bool:
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[404]左叶子之和.py‎

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# 给定二叉树的根节点 root ,返回所有左叶子之和。
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#
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#
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#
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# 示例 1:
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#
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#
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#
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#
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# 输入: root = [3,9,20,null,null,15,7]
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# 输出: 24
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# 解释: 在这个二叉树中,有两个左叶子,分别是 9 和 15,所以返回 24
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#
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#
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# 示例 2:
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#
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#
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# 输入: root = [1]
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# 输出: 0
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#
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#
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#
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#
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# 提示:
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#
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#
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# 节点数在 [1, 1000] 范围内
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# -1000 <= Node.val <= 1000
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#
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#
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#
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 479 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
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total = 0
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def dfs(root):
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nonlocal total
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if not root:
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return
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if root.left and not root.left.left and not root.left.right:
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total+=root.left.val
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dfs(root.left)
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dfs(root.right)
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dfs(root)
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return total
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# leetcode submit region end(Prohibit modification and deletion)
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# 给定一个二叉搜索树的根节点 root 和一个值 key,删除二叉搜索树中的 key 对应的节点,并保证二叉搜索树的性质不变。返回二叉搜索树(有可能被更新)的
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# 根节点的引用。
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#
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# 一般来说,删除节点可分为两个步骤:
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#
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#
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# 首先找到需要删除的节点;
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# 如果找到了,删除它。
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#
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#
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#
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#
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# 示例 1:
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#
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#
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#
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#
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# 输入:root = [5,3,6,2,4,null,7], key = 3
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# 输出:[5,4,6,2,null,null,7]
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# 解释:给定需要删除的节点值是 3,所以我们首先找到 3 这个节点,然后删除它。
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# 一个正确的答案是 [5,4,6,2,null,null,7], 如下图所示。
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# 另一个正确答案是 [5,2,6,null,4,null,7]。
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#
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#
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#
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#
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# 示例 2:
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#
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#
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# 输入: root = [5,3,6,2,4,null,7], key = 0
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# 输出: [5,3,6,2,4,null,7]
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# 解释: 二叉树不包含值为 0 的节点
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#
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#
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# 示例 3:
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#
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#
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# 输入: root = [], key = 0
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# 输出: []
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#
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#
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#
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# 提示:
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#
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#
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# 节点数的范围 [0, 104].
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# -105 <= Node.val <= 105
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# 节点值唯一
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# root 是合法的二叉搜索树
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# -105 <= key <= 105
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#
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#
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#
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#
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# 进阶: 要求算法时间复杂度为 O(h),h 为树的高度。
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# Related Topics 树 二叉搜索树 二叉树 👍 907 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
63+
# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
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# leetcode submit region end(Prohibit modification and deletion)
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# 给你一个含重复值的二叉搜索树(BST)的根节点 root ,找出并返回 BST 中的所有 众数(即,出现频率最高的元素)。
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#
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# 如果树中有不止一个众数,可以按 任意顺序 返回。
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#
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# 假定 BST 满足如下定义:
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#
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#
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# 结点左子树中所含节点的值 小于等于 当前节点的值
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# 结点右子树中所含节点的值 大于等于 当前节点的值
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# 左子树和右子树都是二叉搜索树
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#
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入:root = [1,null,2,2]
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# 输出:[2]
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#
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#
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# 示例 2:
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#
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#
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# 输入:root = [0]
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# 输出:[0]
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#
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#
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#
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#
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# 提示:
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#
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#
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# 树中节点的数目在范围 [1, 104] 内
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# -105 <= Node.val <= 105
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#
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#
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#
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#
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# 进阶:你可以不使用额外的空间吗?(假设由递归产生的隐式调用栈的开销不被计算在内)
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# Related Topics 树 深度优先搜索 二叉搜索树 二叉树 👍 486 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def findMode(self, root: TreeNode) -> List[int]:
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counter = 0
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maxcount = 0
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mode = []
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pre = None
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def inorder(root):
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nonlocal pre, mode, counter, maxcount
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if not root:
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return
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inorder(root.left)
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if pre is None:
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pre = root.val
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counter=1
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elif root.val==pre:
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counter+=1
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else:
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counter=1
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pre = root.val
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if counter > maxcount:
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mode = [root.val]
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maxcount = counter
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elif counter == maxcount:
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mode.append(root.val)
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inorder(root.right)
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inorder(root)
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return mode
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[513]找树左下角的值.py‎

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# 给定一个二叉树的 根节点 root,请找出该二叉树的 最底层 最左边 节点的值。
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#
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# 假设二叉树中至少有一个节点。
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#
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#
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#
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# 示例 1:
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#
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#
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#
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#
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# 输入: root = [2,1,3]
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# 输出: 1
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#
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#
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# 示例 2:
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#
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#
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#
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#
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# 输入: [1,2,3,4,null,5,6,null,null,7]
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# 输出: 7
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#
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#
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#
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#
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# 提示:
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#
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#
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# 二叉树的节点个数的范围是 [1,104]
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# -231 <= Node.val <= 231 - 1
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#
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 358 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
40+
# def __init__(self, val=0, left=None, right=None):
41+
# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
45+
def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
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left = None
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stack = [root]
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while stack:
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l = len(stack)
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for i in range(l):
51+
node = stack.pop(0)
52+
if node.left:
53+
stack.append(node.left)
54+
if node.right:
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stack.append(node.right)
56+
if i==0:
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left = node
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return left.val
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# leetcode submit region end(Prohibit modification and deletion)

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