|
| 1 | + |
| 2 | +>仰望星空的人,不应该被嘲笑 |
| 3 | + |
| 4 | +## 题目描述 |
| 5 | +给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。 |
| 6 | + |
| 7 | +说明: 叶子节点是指没有子节点的节点。 |
| 8 | + |
| 9 | +示例: |
| 10 | + |
| 11 | +```javascript |
| 12 | +给定如下二叉树,以及目标和 sum = 22, |
| 13 | + |
| 14 | + 5 |
| 15 | + / \ |
| 16 | + 4 8 |
| 17 | + / / \ |
| 18 | + 11 13 4 |
| 19 | + / \ / \ |
| 20 | + 7 2 5 1 |
| 21 | +``` |
| 22 | + |
| 23 | +返回: |
| 24 | + |
| 25 | +```javascript |
| 26 | +[ |
| 27 | + [5,4,11,2], |
| 28 | + [5,8,4,5] |
| 29 | +] |
| 30 | +``` |
| 31 | + |
| 32 | +来源:力扣(LeetCode) |
| 33 | +链接:https://leetcode-cn.com/problems/path-sum-ii |
| 34 | +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。 |
| 35 | + |
| 36 | +## 解题思路 |
| 37 | +`dfs`,进行深度优先遍历,一直遍历到子节点为止,进行一次判断,如果当前 `sum`为 0 ,那么就是我们想要的结果,然后注意 `js` 语法中形参如果是数组,那么我们拿到的是引用值,可以拷贝一份。 |
| 38 | + |
| 39 | +```javascript |
| 40 | +/** |
| 41 | + * Definition for a binary tree node. |
| 42 | + * function TreeNode(val) { |
| 43 | + * this.val = val; |
| 44 | + * this.left = this.right = null; |
| 45 | + * } |
| 46 | + */ |
| 47 | +/** |
| 48 | + * @param {TreeNode} root |
| 49 | + * @param {number} sum |
| 50 | + * @return {number[][]} |
| 51 | + */ |
| 52 | +var pathSum = function (root, sum) { |
| 53 | + if(!root) return []; |
| 54 | + let res = []; |
| 55 | + let dfs = (cur, root, sum) => { |
| 56 | + if (root == null) return 0; |
| 57 | + // 拷贝一份 |
| 58 | + cur = [...cur,root.val]; |
| 59 | + sum -= root.val; |
| 60 | + if (!root.left && !root.right && sum == 0) { |
| 61 | + res.push(cur); |
| 62 | + return; |
| 63 | + } |
| 64 | + // 优先遍历左子树 |
| 65 | + root.left && dfs(cur, root.left, sum); |
| 66 | + root.right && dfs(cur, root.right, sum); |
| 67 | + } |
| 68 | + dfs([], root, sum); |
| 69 | + return res; |
| 70 | +}; |
| 71 | +``` |
| 72 | + |
| 73 | +不太明白的小伙伴,这里给一个友好的提示,我们可以打印一下拷贝出来的`cur`,结合图示应该就好理解了,经典的 `dfs`实现的先序遍历。 |
| 74 | + |
| 75 | +```javascript |
| 76 | + 5 |
| 77 | + / \ |
| 78 | + 4 8 |
| 79 | + / / \ |
| 80 | + 11 13 4 |
| 81 | + / \ / \ |
| 82 | + 7 2 5 1 |
| 83 | +``` |
| 84 | + |
| 85 | +```javascript |
| 86 | +[ 5 ] |
| 87 | +[ 5, 4 ] |
| 88 | +[ 5, 4, 11 ] |
| 89 | +[ 5, 4, 11, 7 ] |
| 90 | +[ 5, 4, 11, 2 ] |
| 91 | +[ 5, 8 ] |
| 92 | +[ 5, 8, 13 ] |
| 93 | +[ 5, 8, 4 ] |
| 94 | +[ 5, 8, 4, 5 ] |
| 95 | +[ 5, 8, 4, 1 ] |
| 96 | +``` |
| 97 | + |
| 98 | +## 最后 |
| 99 | +文章产出不易,还望各位小伙伴们支持一波! |
| 100 | + |
| 101 | +往期精选: |
| 102 | + |
| 103 | +<a href="https://github.com/Chocolate1999/Front-end-learning-to-organize-notes">小狮子前端の笔记仓库</a> |
| 104 | + |
| 105 | +<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> |
| 106 | + |
| 107 | +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! |
| 108 | + |
| 109 | + |
| 110 | +<a href="https://yangchaoyi.vip/">访问超逸の博客</a>,方便小伙伴阅读玩耍~ |
| 111 | + |
| 112 | + |
| 113 | + |
| 114 | +```javascript |
| 115 | +学如逆水行舟,不进则退 |
| 116 | +``` |
| 117 | + |
| 118 | + |
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