LeetCode: Path Sum [112]

【题目】

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.

【题意】

判断二叉树中是否存在一条从根到叶子节点的路径，使得路径上的节点值之和等于所给的值

【思路】

    DFS，递归

【代码】

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:        bool dfs(TreeNode*root, int pathSum, int sum){        pathSum+=root->val;        if(root->left==NULL && root->right==NULL){            return pathSum==sum;        }                //搜索左子树        bool hasLeft=false;        if(root->left)            hasLeft=dfs(root->left, pathSum, sum);        //搜索右子树        bool hasRight=false;        if(root->right)            hasRight=dfs(root->right, pathSum, sum);                    return hasLeft || hasRight;            }    bool hasPathSum(TreeNode *root, int sum) {        if(root==NULL)return false;        int pathSum=0;        return dfs(root, pathSum, sum);    }};