Description

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Examples

Example 1:
>Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Example 2:

Example 3:

Constraints:

思路

递归

代码

/**
 * 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 boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null)
            return false;
        if (root.left == null && root.right == null)
            return root.val == targetSum;
        
        boolean canLeft = false, canRight = false;
        if (root.left != null)
            canLeft = hasPathSum(root.left, targetSum - root.val);
        if (canLeft)
            return true;
        if (root.right != null)
            canRight = hasPathSum(root.right, targetSum - root.val);
        
        return canRight;
    }
}