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112. 路径总和

English Version

题目描述

给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。

说明: 叶子节点是指没有子节点的节点。

示例: 
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。

解法

递归求解，递归地询问它的子节点是否能满足条件即可。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        def dfs(root, sum):
            if root is None:
                return False
            if root.val == sum and root.left is None and root.right is None:
                return True
            return dfs(root.left, sum - root.val) or dfs(root.right, sum - root.val)
        return dfs(root, sum)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        return dfs(root, sum);
    }

    private boolean dfs(TreeNode root, int sum) {
        if (root == null) return false;
        if (root.val == sum && root.left == null && root.right == null) return true;
        return dfs(root.left, sum - root.val) || dfs(root.right, sum - root.val);
    }
}

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