Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 

给定一棵树和一个数,判断这个数是不是从根节点到叶子节点的和。

 

递归实现很简单。

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