LeetCode 0112 Path Sum 【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.

Note: A leaf is a node with no children.

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.

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题意

路径上点权值之和是否等于指定的值

思路1

代码1

/**
 * 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;
        int flag = 0;
        dfs(root, 0, sum, flag);
        return flag;
    }
    
    void dfs(TreeNode *root, int s, int sum, int &flag){
        if(flag) return;  // 找到一条路径或者为空
        if(!root) return;
        if(!root->left && !root->right && (s + root->val == sum))
        {
            flag = 1;
            return;
        }
        dfs(root->left, s + root->val, sum, flag);
        dfs(root->right, s + root->val, sum, flag);
    }
};