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: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<TreeNode> s = new Stack<TreeNode>(); 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; } }
原文地址:http://blog.csdn.net/mlweixiao/article/details/38931663