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.
思路：使用先序遍历，不过本人在压栈的时候改变了节点的值。
/**
* Definition for binary tree
* 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;
Stack
s = new Stack
(); s.push(root); while (!s.isEmpty()) { TreeNode temp = s.peek(); s.pop(); if (temp.right != null) { temp.right.val += temp.val; s.push(temp.right); } if (temp.left != null) { temp.left.val += temp.val; s.push(temp.left); } if (temp.left == null && temp.right == null && temp.val == sum) return true; } return false; } }