2、问题分析 对于每个节点,如果其左子节点是叶子,则加上它的值,如果不是,递归,再对右子节点递归即可。 3、代码 1intsumOfLeftLeaves(TreeNode*root) {2if(root ==NULL)3return0;4intans =0;5if(root->left !=NULL) {6if(root->left->left == NULL && root->left->right ==NULL)7ans += roo...
Can you solve this real interview question? Sum of Left Leaves - Given the root of a binary tree, return the sum of all left leaves. A leaf is a node with no children. A left leaf is a leaf that is the left child of another node. Example 1: [https:
classSolution(object):defsumOfLeftLeaves(self, root):ifnotroot:return0ifroot.leftandnotroot.left.leftandnotroot.left.right:returnroot.left.val +self.sumOfLeftLeaves(root.right)returnself.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right) 迭代解法: #Definition for a binary tree no...
LeetCode-404. Sum of Left Leaves Find the sum of all left leaves in a given binary tree. Example: 题目:将一棵树上所有的左叶子节点进行求和。 思路:关键在于找到终止条件,如果一棵树为空,返回0;如果左子树存在且为叶子结点,则将其值加入到sum中去,然后再递归去求左子树中左叶子节点之和和右子树...
【LeetCode】Sum of Left Leaves 左叶子之和 问题Find the sum of all left leaves in a given binary tree. 给定一棵二叉树,找出其所有的左叶子节点的值的和。 Example: 思考 这题的关键在于如何找出所有的左叶子节点。而对于一棵二叉树来说,想要找到其所有的左叶子节点,我想到的最直接的方式是遍历这棵...
404. Sum of Left Leaves Easy Find the sum of all left leaves in a given binary tree. Example: 3 /\ 920 /\ 157 Thereare two left leavesinthe binary tree,withvalues9and15respectively.Return24. packageleetcode.easy; /** * Definition for a binary tree node. public class TreeNode { int...
int left = 0, right = 0; if (head.left != null && head.left.left == null && head.left.right == null) { left = head.left.val; } else { left = sumOfLeftLeaves(head.left); } right = sumOfLeftLeaves(head.right); return left + right; ...
Leetcode: Sum of Left Leaves,Recursion:是不是left子数完全由bottom往上第二层决定,如果是left子树且是叶子节点,那么就是leftleaves,parent得告诉child是不是在left子树BFS:
LeetCode 404 sum of left leaves Find the sum of all left leaves in a given binary tree think about this, when we reached leaf, how are we gonna to know if it is left leaf or not? of course we can modify the iterate version of preorder traverse, it’s very simple....
21 changes: 21 additions & 0 deletions 21 Solutions/404. Sum of Left Leaves.py Original file line numberDiff line numberDiff line change @@ -0,0 +1,21 @@ from typing import Optionalclass TreeNode: def __init__(self, val=0, left=None, right=None):...