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左叶子之和

时间:2024-12-03 09:58:25浏览次数:8  
标签:right TreeNode val 叶子 null root left

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

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