|
| 1 | + |
| 2 | +>仰望星空的人,不应该被嘲笑 |
| 3 | + |
| 4 | +## 题目描述 |
| 5 | + |
| 6 | +给定一个二叉树,找出其最大深度。 |
| 7 | + |
| 8 | +二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。 |
| 9 | + |
| 10 | +说明: 叶子节点是指没有子节点的节点。 |
| 11 | + |
| 12 | +示例: |
| 13 | + |
| 14 | +```javascript |
| 15 | +给定二叉树 [3,9,20,null,null,15,7], |
| 16 | + |
| 17 | + 3 |
| 18 | + / \ |
| 19 | + 9 20 |
| 20 | + / \ |
| 21 | + 15 7 |
| 22 | +``` |
| 23 | + |
| 24 | +返回它的最大深度 3 。 |
| 25 | + |
| 26 | +来源:力扣(LeetCode) |
| 27 | +链接:https://leetcode-cn.com/problems/maximum-depth-of-binary-tree |
| 28 | +著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 |
| 29 | + |
| 30 | + |
| 31 | + |
| 32 | +## 解题思路 |
| 33 | + |
| 34 | +`dfs`,通过后续遍历求得,但是需要注意最后结果要加上根节点(即加1),并且需要判断一下树是不是为空,树为空的话直接返回 `0` 即可,不加 `1` 。 |
| 35 | + |
| 36 | +```javascript |
| 37 | +/** |
| 38 | + * Definition for a binary tree node. |
| 39 | + * function TreeNode(val) { |
| 40 | + * this.val = val; |
| 41 | + * this.left = this.right = null; |
| 42 | + * } |
| 43 | + */ |
| 44 | +/** |
| 45 | + * @param {TreeNode} root |
| 46 | + * @return {number} |
| 47 | + */ |
| 48 | +var maxDepth = function (root) { |
| 49 | + if(!root) return 0; |
| 50 | + let dfs = (root) => { |
| 51 | + if (root == null) return 0; |
| 52 | + // 后续遍历 |
| 53 | + let left = root.left && dfs(root.left) + 1; |
| 54 | + let right = root.right && dfs(root.right) + 1; |
| 55 | + |
| 56 | + return Math.max(left, right); |
| 57 | + } |
| 58 | + // 后续遍历结果还要加上根节点(即加1) |
| 59 | + return dfs(root) + 1; |
| 60 | +}; |
| 61 | +``` |
| 62 | + |
| 63 | + |
| 64 | +**解法二** |
| 65 | + |
| 66 | +`bfs`,一层一层访问,每加一层计数器就加1,这样到达最后一层了直接返回我们的计数器结果即可。 |
| 67 | + |
| 68 | +```javascript |
| 69 | +/** |
| 70 | + * Definition for a binary tree node. |
| 71 | + * function TreeNode(val) { |
| 72 | + * this.val = val; |
| 73 | + * this.left = this.right = null; |
| 74 | + * } |
| 75 | + */ |
| 76 | +/** |
| 77 | + * @param {TreeNode} root |
| 78 | + * @return {number} |
| 79 | + */ |
| 80 | +var maxDepth = function(root) { |
| 81 | + if(!root) return 0 |
| 82 | + let queue =[root] |
| 83 | + let cnt = 0 |
| 84 | + while(queue.length){ |
| 85 | + let size = queue.length |
| 86 | + while(size--){ |
| 87 | + let node = queue.shift() |
| 88 | + if(node.left) queue.push(node.left) |
| 89 | + if(node.right) queue.push(node.right) |
| 90 | + } |
| 91 | + ++cnt |
| 92 | + } |
| 93 | + return cnt |
| 94 | +}; |
| 95 | +``` |
| 96 | + |
| 97 | + |
| 98 | + |
| 99 | + |
| 100 | +## 最后 |
| 101 | +文章产出不易,还望各位小伙伴们支持一波! |
| 102 | + |
| 103 | +往期精选: |
| 104 | + |
| 105 | +<a href="https://github.com/Chocolate1999/Front-end-learning-to-organize-notes">小狮子前端の笔记仓库</a> |
| 106 | + |
| 107 | +<a href="https://github.com/Chocolate1999/leetcode-javascript">leetcode-javascript:LeetCode 力扣的 JavaScript 解题仓库,前端刷题路线(思维导图)</a> |
| 108 | + |
| 109 | +小伙伴们可以在Issues中提交自己的解题代码,🤝 欢迎Contributing,可打卡刷题,Give a ⭐️ if this project helped you! |
| 110 | + |
| 111 | + |
| 112 | +<a href="https://yangchaoyi.vip/">访问超逸の博客</a>,方便小伙伴阅读玩耍~ |
| 113 | + |
| 114 | + |
| 115 | + |
| 116 | +```javascript |
| 117 | +学如逆水行舟,不进则退 |
| 118 | +``` |
| 119 | + |
| 120 | + |
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