Path Sum问题及解法

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问题描述：

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.

示例：

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.

问题分析：

求某一路径上的和是否为给定的数。只需要每次遍历到根节点就计算出当前剩余值是否根结点的值相等即可，主要原理还是利用深度优先遍历。

过程详见代码：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool hasPathSum(TreeNode* root, int sum) {        if(root == NULL) return false;        if(root->left == NULL && root->right == NULL && sum == root->val) return true;// 判断是否为叶子节点         return hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val);    }};

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