112. Path Sum

目录

• 题目描述
• 代码实现

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.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

代码实现

class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
      //注意这里递归终止条件的层层递进的设计方法
    	if(root == nullptr)
    		return false;
    	int curSum = sum - root->val;  
    	int result = false; 
      //到达叶节点，并且满足给定条件  
    	if(curSum == 0 && root->left == nullptr && root->right == nullptr)
    		return true;
    	if(!result)
    		result = hasPathSum(root->left,curSum);
    	if(!result)
    		result = hasPathSum(root->right,curSum);
    	return result;    
    }
};