给定二叉树的根节点 root
,返回所有左叶子之和。
示例 1:
输入: root = [3,9,20,null,null,15,7] 输出: 24 解释: 在这个二叉树中,有两个左叶子,分别是 9 和 15,所以返回 24
示例 2:
输入: root = [1] 输出: 0
提示:
- 节点数在
[1, 1000]
范围内 -1000 <= Node.val <= 1000
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
int result = 0;
if (root != null) {
if (root.left != null) {
if (root.left.left == null && root.left.right == null) {
result = root.left.val;
}
result += sumOfLeftLeaves(root.left);
}
if (root.right != null) {
result += sumOfLeftLeaves(root.right);
}
}
return result;
}
}
标签:right,TreeNode,val,叶子,null,root,left
From: https://blog.csdn.net/linsa_pursuer/article/details/144181264