Binary Tree Inorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

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

Iteratively:

 1 /**
 2  * Definition for binary tree
 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         TreeNode *sneil = NULL;
14         vector<int> rst;
15         if(root==NULL)
16             return rst;
17             
18         stack<TreeNode*> istack;
19         istack.push(sneil);
20         while(!istack.empty()){
21             while(root->left){
22                 istack.push(root);
23                 root = root->left;
24             }
25             while(root!=NULL && root->right==NULL){
26                 rst.push_back(root->val);
27                 root = istack.top();
28                 istack.pop();
29             }
30             if(root==NULL)
31                 break;
32             rst.push_back(root->val);
33             root = root->right;
34         }
35         return rst;
36     }
37 };

不使用stack的解法，明天再写。。

转载于:https://www.cnblogs.com/nnoth/p/3744111.html

标签：Binary,TreeNode,root,Tree,istack,rst,return,NULL,Inorder
来源： https://blog.csdn.net/weixin_33725270/article/details/93847194