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yanglbmeyanglbme
committedMar 28, 2021
feat: update leetcode solutions: No.0404. Sum of Left Leaves
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‎README.md

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- [二分查找](/basic/searching/BinarySearch/README.md)
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- [二分查找 II](/basic/searching/BinarySearch-II/README.md)
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## 面试高频考题
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## 高频考题
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### 数组
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‎index.html

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<head>
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<meta charset="UTF-8">
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<title>LeetCode、剑指Offer、程序员面试金典题解</title>
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<title>LeetCode & Coding Interview Guide</title>
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<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" />
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<meta name="keywords"
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content="doc,docs,doocs,documentation,github,gitee,coding,pages,leetcode,coding-interview,cracking-the-coding-interview,yanglbme">
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content="doc,docs,doocs,documentation,github,gitee,coding,pages,leetcode,coding-interview,coding-interview-guide,cracking-the-coding-interview,yanglbme">
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<meta name="description" content="LeetCode、剑指Offer、程序员面试金典题解">
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<meta name="viewport"
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content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
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</head>
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<body>
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<div id="app">本系列知识由 Doocs 技术社区原创发布</div>
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<div id="app">LeetCode & Coding Interview Guide</div>
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<script>
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window.$docsify = {
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name: 'leetcode',

‎solution/0400-0499/0404.Sum of Left Leaves/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def sumOfLeftLeaves(self, root: TreeNode) -> int:
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if root is None:
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return 0
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res = 0
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if root.left and root.left.left is None and root.left.right is None:
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res += root.left.val
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res += self.sumOfLeftLeaves(root.left)
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res += self.sumOfLeftLeaves(root.right)
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return res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public int sumOfLeftLeaves(TreeNode root) {
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if (root == null) {
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return 0;
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}
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int res = 0;
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if (root.left != null && root.left.left == null && root.left.right == null) {
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res += root.left.val;
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}
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res += sumOfLeftLeaves(root.left);
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res += sumOfLeftLeaves(root.right);
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return res;
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}
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}
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```
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### **...**

‎solution/0400-0499/0404.Sum of Left Leaves/README_EN.md

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### **Python3**
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def sumOfLeftLeaves(self, root: TreeNode) -> int:
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if root is None:
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return 0
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res = 0
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if root.left and root.left.left is None and root.left.right is None:
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res += root.left.val
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res += self.sumOfLeftLeaves(root.left)
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res += self.sumOfLeftLeaves(root.right)
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return res
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```
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### **Java**
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public int sumOfLeftLeaves(TreeNode root) {
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if (root == null) {
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return 0;
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}
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int res = 0;
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if (root.left != null && root.left.left == null && root.left.right == null) {
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res += root.left.val;
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}
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res += sumOfLeftLeaves(root.left);
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res += sumOfLeftLeaves(root.right);
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return res;
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}
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}
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```
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### **...**

‎solution/0400-0499/0404.Sum of Left Leaves/Solution.java

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* }
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*/
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class Solution {
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private int sum = 0;
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public int sumOfLeftLeaves(TreeNode root) {
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if (root == null) return 0;
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if (root == null) {
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return 0;
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}
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int res = 0;
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if (root.left != null && root.left.left == null && root.left.right == null) {
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sum += root.left.val;
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res += root.left.val;
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}
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sumOfLeftLeaves(root.left);
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sumOfLeftLeaves(root.right);
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return sum;
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res += sumOfLeftLeaves(root.left);
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res += sumOfLeftLeaves(root.right);
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return res;
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}
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}

‎solution/0400-0499/0404.Sum of Left Leaves/Solution.py

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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def sumOfLeftLeaves(self, root: TreeNode) -> int:
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if root is None:
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return 0
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res = 0
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if root.left and root.left.left is None and root.left.right is None:
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res += root.left.val
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res += self.sumOfLeftLeaves(root.left)
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res += self.sumOfLeftLeaves(root.right)
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return res

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