Leetcode 112 Path Sum

最新推荐文章于 2020-06-08 12:39:15 发布

转载最新推荐文章于 2020-06-08 12:39:15 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/qinduanyinghua/p/6544564.html

本文介绍了一种算法，用于判断给定的二叉树是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的数值。通过递归和迭代两种方法实现，适用于算法学习和面试准备。

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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.

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode* root, int sum) {
13         if(root == NULL)
14             return false;
15         if(root->left == NULL && root->right == NULL && sum == root->val)
16             return true;
17         if(hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val))
18             return true;
19         else
20             return false;
21     }
22 };

 1 bool hasPathSum(TreeNode* root, int sum){
 2         if(root == NULL)
 3             return false;
 4         queue<TreeNode*> q;
 5         q.push(root);
 6         TreeNode* temp;
 7         while(!q.empty()){
 8             int size = q.size();
 9             while(size--){
10                 temp = q.front();
11                 q.pop();
12                 if(temp->left != NULL){
13                     q.push(temp->left);
14                     temp->left->val += temp->val;
15                 }
16                 if(temp->right != NULL){
17                     q.push(temp->right);
18                     temp->right->val += temp->val;
19                 }
20                 if(temp->left == NULL && temp->right == NULL && (temp->val == sum)){
21                     return true;
22                 }
23             }
24         }
25         return false;
26     }

转载于:https://www.cnblogs.com/qinduanyinghua/p/6544564.html