[LeetCode] Path Sum

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.

     这道题挺简单的，直接用recursive做就可以了。因为反正每一个node检查的方式都一样。

     只要目前的sum减去正在检查的node的value之后等于0，就可以return true。

     所以程序还是很简单的。

/**
 * 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;
        }
        sum=sum-root.val;
        if(root.left==null&&root.right==null){
            if(sum==0){
                return true;
            }
            return false;
        }
        return hasPathSum(root.left,sum)||hasPathSum(root.right,sum);
    }
}

[LeetCode] Path Sum

原文：http://www.cnblogs.com/orangeme404/p/4744191.html