Binary Tree Inorder Traversal

题目：

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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

解析：

 递归：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> v;
        inTraversal(root,v);
        return v;
    }
    
    void inTraversal(TreeNode* root,vector<int>& v)
    {
        if(!root) return;
        inTraversal(root->left,v);
        v.push_back(root->val);
        inTraversal(root->right,v);
    }
};

 栈非递归

class Solution
{
public:
    vector<int> inorderTraversal(TreeNode* root)
    {
        vector<int> v;
        if(!root) return v;
        TreeNode* temp = root;
        stack<TreeNode*> s;
        while(true)
        {
            while(temp)
            {
                s.push(temp);
                temp = temp->left;
            }
            if(s.empty()) break;
            temp = s.top();
            s.pop();
            v.push_back(temp->val);
            temp = temp->right;
        }
        return v;
    }
};

不用栈：

// iterate, morris traversal, without stack
vector<int> inorderTraversal(TreeNode* root) {
    vector<int> v;
    if(!root) return v;
    TreeNode* temp = root, *prev;
    while(temp){
        if(!temp->left){
            v.push_back(temp->val);
            temp = temp->right;
        }else{
            prev = temp->left;
            while(prev->right&&(prev->right != temp))
                prev = prev->right;
            if(!prev->right){
                prev->right = temp;
                temp = temp->left;
            }else{
                v.push_back(temp->val);
                prev->right = NULL;
                temp = temp->right;
            }
        }
    }
}