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