LeetCode: 112. Path Sum

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 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.

题目大意: 判断给定二叉树中是否有个根节点到叶节点的路径，使得路径上的数值的和是给定的数。

解题思路 —— 递归求解

要满足从 root 到叶节点的和为 sum， 只需要满足从左/右孩子到叶节点的和为 sum - root->val 即可。

AC 代码

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left;[](h * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if(root == nullptr) return false;
        if(root->left == nullptr && root->right == nullptr && root->val == sum) return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }

实际上，可以合并成一行代码:

/** * 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) {
        return root != nullptr && 
        ((root->left == nullptr && root->right == nullptr && root->val == sum) || 
        hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val));
    }
};