inorderTraversal.cpp

/*
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].

   1
  / \
 2   3
    /
   4
    \
     5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
*/


/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> nodes;
        TreeNode* tmp;
        if(root==nullptr)
            return result;
        nodes.push(root);
        while(!nodes.empty())
        {
            tmp=nodes.top();
            if(tmp->left)
            {
                 nodes.push(tmp->left);
                 tmp->left=nullptr;
            }else
            {
               result.push_back(tmp->val);
               nodes.pop();
               if(tmp->right)
                  nodes.push(tmp->right);
            }
        }
        return result;
    }
};




//better Solution,this will not changed the node.
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> vector;
        if(!root)
        return vector;
        unordered_map<TreeNode *, bool> map;//left child has been visited:true.
        stack<TreeNode *> stack;
        stack.push(root);
        while(!stack.empty())
        {
            TreeNode *pNode = stack.top();
            if(pNode->left && !map[pNode])
            {
                stack.push(pNode->left);
                map[pNode] = true;
            }
            else
            {
                vector.push_back(pNode->val);
                stack.pop();
                if(pNode->right)
                    stack.push(pNode->right);
            }
        }
        return vector;
    }
};



//Best Solution,using one stack and will not changed the node. 
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> vector;
        stack<TreeNode *> stack;
        TreeNode *pCurrent = root;

        while(!stack.empty() || pCurrent)
        {
            if(pCurrent)
            {
                stack.push(pCurrent);
                pCurrent = pCurrent->left;
            }
            else
            {
                TreeNode *pNode = stack.top();
                vector.push_back(pNode->val);
                stack.pop();
                pCurrent = pNode->right;
            }
        }
        return vector;
    }
};