小小程序媛  
得之坦然,失之淡然,顺其自然,争其必然

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,
1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

分析

判断给定树中有无 【跟—>叶子】路径节点值之和为 给定sum

递归实现思想:

  1. 空树,返回false
  2. 单根节点,判断,若等于sum返回true,否则返回false
  3. 更新sum -= root—>val 递归判断左右子树,其中一真即真。

AC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) {
        //空树返回false
        if (root == NULL)
        {
            return false;
        }
        //若是叶子节点,判断返回
        else if (!root->left && !root->right)
        {
            if (root->val == sum)
                return true;
            else
                return false;
        }
        else
        {
            //否则,递归判断左右子树
            sum -= root->val;

            return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
        }
    }
};

GitHub测试程序源码

posted on 2015-10-19 12:53  Coding菌  阅读(105)  评论(0编辑  收藏  举报