Path Sum

标签：二叉树

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.

#include<iostream>
#include<vector>
using namespace std;

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
bool hasPathSum(TreeNode *root, int sum) {
	if (root == NULL)
		return false;
	return judge(root, sum);
}
bool judge(TreeNode *root, int sum) {
	if (root == NULL)
		return false;
	if (root->left==NULL&&root->right==NULL)
		return sum-root->val==0? true : false;
	return judge(root->left, sum - root->val) || judge(root->right, sum - root->val);
}

Path Sum

标签：二叉树

原文地址：http://blog.csdn.net/li_chihang/article/details/44749113