Description

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.

Explain

這道題一看就是深度優(yōu)先搜索的問題了。題意很簡(jiǎn)單，就是從找一個(gè)條路，從根部到葉子，其中節(jié)點(diǎn)的值加起來(lái)等于給定的值。那就遍歷每個(gè)節(jié)點(diǎn)，每次到達(dá)一個(gè)新的葉子節(jié)點(diǎn)時(shí)，加上當(dāng)前結(jié)點(diǎn)的值，看是否等于給定的值，如果是，那么就找到了。如果不是，退回上一層繼續(xù)遍歷，直到遍歷完整個(gè)樹，如果還找不到，那就無(wú)解。

Solution

/**
 * 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) {
        bool flag = false;
        if (!root) return false;
        dfs(flag, root, sum, 0);
        return flag;      
    }
    void dfs(bool& flag, TreeNode* root, int sum, int _sum) {
        if (flag) return;
        if (root) {
            dfs(flag, root->left, sum,_sum + root->val);
            dfs(flag, root->right, sum,_sum + root->val);
            _sum += root->val;
            if(!root->left&&!root->right&&_sum == sum) flag = true;
        }
    }
};