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Commit 458848d

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leetcode_cpp
Change-Id: I90cac0eb020fb8e1f9ba56d803cfe7f7f8ec2fcd
1 parent 7b54024 commit 458848d

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‎cpp/leetcode/101.对称二叉树.cpp

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/*
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* @lc app=leetcode.cn id=101 lang=cpp
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*
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* [101] 对称二叉树
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*
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* https://leetcode-cn.com/problems/symmetric-tree/description/
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*
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* algorithms
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* Easy (55.21%)
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* Likes: 1412
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* Dislikes: 0
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* Total Accepted: 337.9K
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* Total Submissions: 611.9K
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* Testcase Example: '[1,2,2,3,4,4,3]'
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*
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* 给定一个二叉树,检查它是否是镜像对称的。
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*
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*
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*
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* 例如,二叉树 [1,2,2,3,4,4,3] 是对称的。
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*
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* ⁠ 1
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* ⁠ / \
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* ⁠ 2 2
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* ⁠/ \ / \
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* 3 4 4 3
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*
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*
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*
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*
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* 但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:
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*
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* ⁠ 1
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* ⁠ / \
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* ⁠ 2 2
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* ⁠ \ \
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* ⁠ 3 3
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*
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*
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*
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*
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* 进阶:
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*
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* 你可以运用递归和迭代两种方法解决这个问题吗?
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*
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*/
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// Definition for a binary tree node.
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struct TreeNode {
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int val;
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TreeNode *left;
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TreeNode *right;
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TreeNode() : val(0), left(nullptr), right(nullptr) {}
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TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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TreeNode(int x, TreeNode *left, TreeNode *right)
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: val(x), left(left), right(right) {}
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};
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// @lc code=start
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class Solution {
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private:
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bool check(TreeNode *left, TreeNode *right) {
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if (left == nullptr && right == nullptr) {
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return true;
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}
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if (left == nullptr || right == nullptr) {
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return false;
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}
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return left->val == right->val && check(left->left, right->right) &&
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check(left->right, right->left);
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}
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public:
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bool isSymmetric(TreeNode *root) {
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return root != nullptr && check(root->left, root->right);
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}
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};
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=102 lang=cpp
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*
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* [102] 二叉树的层序遍历
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*
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* https://leetcode-cn.com/problems/binary-tree-level-order-traversal/description/
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*
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* algorithms
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* Medium (64.14%)
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* Likes: 895
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* Dislikes: 0
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* Total Accepted: 323.6K
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* Total Submissions: 504.5K
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* Testcase Example: '[3,9,20,null,null,15,7]'
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*
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* 给你一个二叉树,请你返回其按 层序遍历 得到的节点值。
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* (即逐层地,从左到右访问所有节点)。
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*
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*
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*
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* 示例:
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* 二叉树:[3,9,20,null,null,15,7],
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*
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*
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* ⁠ 3
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* ⁠ / \
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* ⁠ 9 20
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* ⁠ / \
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* ⁠ 15 7
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*
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*
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* 返回其层序遍历结果:
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*
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*
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* [
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* ⁠ [3],
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* ⁠ [9,20],
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* ⁠ [15,7]
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* ]
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*
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*
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*/
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#include <queue>
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#include <vector>
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// Definition for a binary tree node.
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struct TreeNode {
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int val;
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TreeNode *left;
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TreeNode *right;
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TreeNode() : val(0), left(nullptr), right(nullptr) {}
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TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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TreeNode(int x, TreeNode *left, TreeNode *right)
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: val(x), left(left), right(right) {}
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};
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// @lc code=start
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class Solution {
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public:
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std::vector<std::vector<int>> levelOrder(TreeNode *root) {
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std::vector<std::vector<int>> ret;
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if (root == nullptr) {
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return ret;
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}
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std::queue<TreeNode *> q;
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q.push(root);
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while (!q.empty()) {
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int batch = q.size();
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std::vector<int> level(batch);
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for (int i = 0; i < batch; i++) {
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TreeNode *node = q.front();
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q.pop();
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level[i] = node->val;
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if (node->left) {
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q.push(node->left);
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}
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if (node->right) {
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q.push(node->right);
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}
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}
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ret.push_back(level);
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}
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return ret;
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}
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};
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=105 lang=cpp
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*
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* [105] 从前序与中序遍历序列构造二叉树
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*
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* https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/
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*
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* algorithms
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* Medium (71.39%)
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* Likes: 1873
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* Dislikes: 0
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* Total Accepted: 460.6K
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* Total Submissions: 645.1K
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* Testcase Example: '[3,9,20,15,7]\n[9,3,15,20,7]'
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*
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* 给定两个整数数组 preorder 和 inorder ,其中 preorder 是二叉树的先序遍历,
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* inorder 是同一棵树的中序遍历,请构造二叉树并返回其根节点。
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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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* 输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
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* 输出: [3,9,20,null,null,15,7]
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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: preorder = [-1], inorder = [-1]
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* 输出: [-1]
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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 <= 3000
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* inorder.length == preorder.length
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* -3000 <= preorder[i], inorder[i] <= 3000
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* preorder 和 inorder 均 无重复 元素
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* inorder 均出现在 preorder
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* preorder 保证 为二叉树的前序遍历序列
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* inorder 保证 为二叉树的中序遍历序列
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*
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*
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*/
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#include <unordered_map>
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#include <vector>
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// Definition for a binary tree node.
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struct TreeNode {
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int val;
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TreeNode *left;
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TreeNode *right;
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TreeNode() : val(0), left(nullptr), right(nullptr) {}
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TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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TreeNode(int x, TreeNode *left, TreeNode *right)
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: val(x), left(left), right(right) {}
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};
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// @lc code=start
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class Solution {
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private:
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std::unordered_map<int, int> inorder_index;
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TreeNode *buildTree(std::vector<int> &preorder, int pre_left, int pre_right,
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std::vector<int> &inorder, int in_left, int in_right) {
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if (pre_left > pre_right) {
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return nullptr;
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}
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int mid = inorder_index[preorder[pre_left]] - in_left; // count
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TreeNode *root = new TreeNode(preorder[pre_left]);
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root->left = buildTree(preorder, pre_left + 1, pre_left + mid, inorder,
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in_left, in_left + mid - 1);
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root->right = buildTree(preorder, pre_left + mid + 1, pre_right, inorder,
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in_left + mid + 1, in_right);
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return root;
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}
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public:
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TreeNode *buildTree(std::vector<int> &preorder, std::vector<int> &inorder) {
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int n = inorder.size();
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for (int i = 0; i < n; i++) {
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inorder_index[inorder[i]] = i;
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}
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return buildTree(preorder, 0, n - 1, inorder, 0, n - 1);
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}
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};
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=121 lang=cpp
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*
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* [121] 买卖股票的最佳时机
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*
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* https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/description/
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*
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* algorithms
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* Easy (56.84%)
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* Likes: 1664
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* Dislikes: 0
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* Total Accepted: 463K
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* Total Submissions: 814.4K
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* Testcase Example: '[7,1,5,3,6,4]'
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*
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* 给定一个数组 prices ,它的第 i 个元素 prices[i] 表示一支给定股票第 i
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* 天的价格。
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*
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* 你只能选择 某一天 买入这只股票,并选择在 未来的某一个不同的日子
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* 卖出该股票。设计一个算法来计算你所能获取的最大利润。
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*
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* 返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润,返回 0 。
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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:[7,1,5,3,6,4]
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* 输出:5
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* 解释:在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票价格 =
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* 6)的时候卖出,最大利润 = 6-1 = 5 。 ⁠ 注意利润不能是 7-1 = 6,
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* 因为卖出价格需要大于买入价格;同时,你不能在买入前卖出股票。
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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:prices = [7,6,4,3,1]
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* 输出:0
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* 解释:在这种情况下, 没有交易完成, 所以最大利润为 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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* 1
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* 0
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*
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*
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*/
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#include <algorithm>
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#include <vector>
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// @lc code=start
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class Solution {
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public:
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int maxProfit(std::vector<int> &prices) {
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if (prices.empty()) {
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return 0;
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}
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int buy = prices[0];
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int profit = 0;
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for (int i = 1; i < prices.size(); i++) {
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buy = std::min(buy, prices[i]);
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profit = std::max(profit, prices[i] - buy);
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}
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return profit;
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}
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};
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// @lc code=end

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