leetcode:Path Sum (路径之和) 【面试算法题】

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 ,

              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.

题意判断从根节点到叶子节点的路径和是否有等于sum的值。


叶子节点就是没有子节点的节点,判断叶子节点的当前值是否等于当前的sum。

若不是叶子节点,将当前sum减去当前节点的值,递归求解。


/**
 * Definition for binary tree
 * 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)return false;
        if(sum==root->val&&root->left==NULL&&root->right==NULL)return true;
        return (root->left && hasPathSum(root->left,sum-root->val) )
             ||(root->right && hasPathSum(root->right,sum-root->val));
    }
};
// blog.csdn.net/havenoidea


原文地址:https://www.cnblogs.com/fuhaots2009/p/3373732.html