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: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.
思路:dfs到叶子节点的时候再判断和是否满足。
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean dfs(TreeNode root, int sum, int cur) {
		if (root == null) return false;
		if (root.left == null && root.right == null) 
			return sum == (cur + root.val);
		
		return dfs(root.left, sum, cur+root.val) || dfs(root.right, sum, cur+root.val);
	}
	
    public boolean hasPathSum(TreeNode root, int sum) {
    	return dfs(root, sum, 0);
    }
}
原文:http://blog.csdn.net/u011345136/article/details/45600873