94. Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

 

 

非递归二叉树中序遍历

 

java：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public List<Integer> inorderTraversal(TreeNode root) {
12         List<Integer> res = new ArrayList<>() ;
13         if (root == null)
14             return res ;
15         Stack<TreeNode> stack = new Stack<>() ;
16         TreeNode cur = root ;
17         while(cur != null || !stack.isEmpty()){
18             while(cur != null){
19                 stack.push(cur) ;
20                 cur = cur.left ;
21             }
22             TreeNode node = stack.pop() ;
23             res.add(node.val) ;
24             cur  = node.right ;
25         }
26         return res ;
27     }
28 }

 

 

C++:

 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     vector<int> inorderTraversal(TreeNode* root) {
13         vector<int> res ;
14         if (root == NULL)
15             return res ;
16         stack<TreeNode*> s ;
17         TreeNode* cur = root ;
18         while(cur != NULL || !s.empty()){
19             while(cur != NULL){
20                 s.push(cur) ;
21                 cur = cur->left ;
22             }
23             TreeNode* node = s.top() ;
24             s.pop() ;
25             res.push_back(node->val) ;
26             cur = node->right ;
27         }
28         return res ;
29     }
30 };

 

posted @ 2019-01-11 11:22  __Meng  阅读(117)  评论(0)  编辑  收藏  举报