题目描述

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 andsum = 22,

return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

题目大意

给定一个二叉树和一个sum值，看看是否有一条从根到叶子结点的路径，使得这条路径上各个结点的值相加刚好等于sum。

思路

递归思路求解，递归每条路径，看是否有一条路径之和刚好等于sum。

递归出口是：
（1）返回为true：当前结点刚好是叶子结点&&sum-当前结点的值为0。
（2）返回为false：root == NULL。

避坑：sum和各个结点值有可能是负数；

代码

#include<iostream>
using namespace std;

// Definition for binary tree
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

typedef TreeNode* tree;

bool hasPathSum(TreeNode *root, int sum)
{
    if(root == NULL)return false;

    // 保证该店是叶子结点且加到这儿的和为sum
    if(root->val==sum && root->left==NULL && root->right==NULL)
        return true;

    return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}

int main()
{
    tree t = new TreeNode(-2);
    t->left = new TreeNode(-3);
    cout<<hasPathSum(t, -5)<<endl;
    //cout<<-5-(-2)<<endl;
    return 0;
}

以上。