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Commit aa785e3

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add LeetCode 404. 左叶子之和
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![](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9jZG4uanNkZWxpdnIubmV0L2doL2Nob2NvbGF0ZTE5OTkvY2RuL2ltZy8yMDIwMDgyODE0NTUyMS5qcGc?x-oss-process=image/format,png)
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>仰望星空的人,不应该被嘲笑
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## 题目描述
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计算给定二叉树的所有左叶子之和。
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示例:
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```javascript
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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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在这个二叉树中,有两个左叶子,分别是 915,所以返回 24
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```
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来源:力扣(LeetCode)
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链接:https://leetcode-cn.com/problems/sum-of-left-leaves
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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。
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## 解题思路
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`dfs`,求左叶子之和,叶子结点我们比较好判断,而对于左孩子,我们设置一个标记就好了,例如左孩子标记 `1`,右孩子标记 `0`,那么当且仅当是叶子节点,并且标记为 `1`(即左孩子)时,我们进行累加求和。
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```javascript
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {number}
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*/
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var sumOfLeftLeaves = function (root) {
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if (!root) return 0;
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let res = 0;
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let dfs = (cur,root) => {
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// 判断是叶子节点并且是左子树
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if (!root.left && !root.right && cur) {
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res += root.val;
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return;
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}
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// 先遍历左子树在遍历右子树
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// 设置cur为1代表是左子树,为0代表右子树
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root.left && dfs(1,root.left);
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root.right && dfs(0,root.right);
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}
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dfs(0,root);
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return res;
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};
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```
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## 最后
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文章产出不易,还望各位小伙伴们支持一波!
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往期精选:
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<a href="https://github.com/Chocolate1999/Front-end-learning-to-organize-notes">小狮子前端の笔记仓库</a>
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<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>
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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!
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<a href="https://yangchaoyi.vip/">访问超逸の博客</a>,方便小伙伴阅读玩耍~
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![](https://img-blog.csdnimg.cn/2020090211491121.png#pic_center)
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```javascript
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学如逆水行舟,不进则退
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```
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