leetcode 112. Path Sum

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.

Note: A leaf is a node with no children.

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.

递归方法：

 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 private:
12     bool isLeaf(TreeNode *root) {
13         return (root->left == NULL) && (root->right == NULL);
14     }
15 public:
16     bool hasPathSum(TreeNode* root, int sum) {
17         if (root == NULL)
18             return false;
19         if (isLeaf(root) && sum == root->val)
20             return true;
21         return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
22     }
23 };

非递归：

 

 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         queue<TreeNode*> q;
16         q.push(root);
17         TreeNode *temp;
18         while (!q.empty()) {
19             int size = q.size();
20             while (size--) {
21                 temp = q.front();
22                 q.pop();
23                 if (temp->left != NULL) {
24                     q.push(temp->left);
25                     temp->left->val += temp->val;
26                 }
27                 if (temp->right != NULL) {
28                     q.push(temp->right);
29                     temp->right->val += temp->val;
30                 }
31                 if (temp->left == NULL && temp->right == NULL && temp->val == sum) {
32                     return true;
33                 }
34                     
35             }
36         }
37         return false;
38     }
39 };

 

posted @ 2019-08-29 23:03  琴影  阅读(213)  评论(0)  编辑  收藏  举报