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.
C++版
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum) {
if (root == NULL) return false;
if (root->left == NULL && root->right == NULL && sum - root->val == 0) return true;
return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}
};
Java版本：
/**
* 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;
if (root.left == null && root.right == null && sum - root.val == 0) return true;
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
原文：http://www.cnblogs.com/zlz-ling/p/4036689.html