Binary Tree Postorder Traversal Solutions in C++

// Source : https://oj.leetcode.com/problems/binary-tree-postorder-traversal/
// Author : Hao Chen
// Date   : 2014-07-21



#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

vector<int> postorderTraversal1(TreeNode *root);
vector<int> postorderTraversal2(TreeNode *root);


// We have two methods here.
//   1) the first one acutally is pre-order but reversed the access order.
//   2) the second one is the traditional one 

vector<int> postorderTraversal(TreeNode *root) {
    if (random()%2){
        cout << "---method one---" << endl;
        return postorderTraversal1(root);
    }
    cout << "---method two---" << endl;
    return postorderTraversal2(root);
}


vector<int> postorderTraversal1(TreeNode *root) {
    vector<int> v;
    vector<TreeNode*> stack;
    if (root) {
        stack.push_back(root);
    }
    while (stack.size()>0){
        TreeNode *n = stack.back();
        stack.pop_back();
        v.push_back(n->val);
        if (n->left){
            stack.push_back(n->left);
        } 
        if (n->right) {
            stack.push_back(n->right);
        }
    }
    std::reverse(v.begin(), v.end());  // the trick
    return v;
}

// traditional and standard way.
// using the stack to simulate the recursive function stack.
vector<int> postorderTraversal2(TreeNode *root) {
    vector<int> v;
    vector<TreeNode*> stack;
    TreeNode *node = root;
    TreeNode *lastVisitNode = NULL;
    while(stack.size()>0 || node!=NULL){

        if (node != NULL){
            // keep going the left
            stack.push_back(node);
            node = node->left;
        }else{
            TreeNode *n = stack.back();
            // left way is finsised, keep going to the right way
            if (n->right != NULL && lastVisitNode != n->right){
                node = n->right;
            }else{
                // both left and right has been accessed.
                stack.pop_back();
                v.push_back(n->val);
                lastVisitNode = n;
            }
        }
    }
    return v;
}


TreeNode* createTree(int a[], int n)
{
    if (n<=0) return NULL;

    TreeNode **tree = new TreeNode*[n];

    for(int i=0; i<n; i++) {
        if (a[i]==0 ){
            tree[i] = NULL;
            continue;
        }
        tree[i] = new TreeNode(a[i]);
    }
    int pos=1;
    for(int i=0; i<n && pos<n; i++) {
        if (tree[i]){
            tree[i]->left = tree[pos++];
            if (pos<n){
                tree[i]->right = tree[pos++];
            }
        }
    }
    return tree[0];
}

void printTree_post_order(TreeNode *root)
{
    if (root == NULL){
        //cout << "# ";
        return ;
    }
    printTree_post_order(root->left);
    printTree_post_order(root->right);
    cout << root->val << " ";
}


void printArray(vector<int> v)
{
    for(int i=0; i<v.size(); i++){
        cout << v[i] << " ";
    }
    cout << endl;
}

int main()
{
    srand(time(0));
    int a[] = {1,2,3,4,5,0,6,0,0,7,8,9,0};
    TreeNode* p = createTree(a, sizeof(a)/sizeof(int));
    printTree_post_order(p);
    cout << endl;
    vector<int> v = postorderTraversal(p);
    printArray(v);
    cout << endl;

    return 0;
}

#include <vector>
#include <stack>
#include <cstddef>
using std::vector; using std::stack;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> retv;
        TreeNode *last = NULL;
        for (stack<TreeNode*> stk; root || !stk.empty(); )
        {
            if (root)
            {
                stk.push(root);
                root = root->left;
            }
            else if (stk.top()->right && stk.top()->right != last)
            {
                root = stk.top()->right;
            }
            else
            {
                last = stk.top();
                retv.push_back(last->val); stk.pop();
            }
        }
        return retv;
    }
};

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2017-11-17

#include <iostream>
#include <vector>

using namespace std;


/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

// Recursive
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        __postorderTraversal(root, res);
        return res;
    }

private:
    void __postorderTraversal(TreeNode* node, vector<int> &res){

        if( node ){
            __postorderTraversal(node->left, res);
            __postorderTraversal(node->right, res);
            res.push_back(node->val);
        }
    }
};

int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2017-11-17

#include <iostream>
#include <vector>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// My Non-Recursive
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
class Solution {

private:
    struct Command{
        string s;   // go, print
        TreeNode* node;
        Command(string s, TreeNode* node): s(s), node(node){}
    };

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        stack<Command> stack;
        stack.push(Command("go", root) );
        while(!stack.empty()){
            Command command = stack.top();
            stack.pop();

            if(command.s == "print")
                res.push_back(command.node->val);
            else{
                assert(command.s == "go");
                stack.push(Command("print", command.node));
                if(command.node->right)
                    stack.push(Command("go",command.node->right));
                if(command.node->left)
                    stack.push(Command("go",command.node->left));
            }
        }
        return res;
    }
};

int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-05-30

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// Non-Recursive
// Using a tag to record whether the node has been visited
//
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
class Solution {

private:
    struct TagNode{
        TreeNode* node;
        bool isFirst;
        TagNode(TreeNode* node): node(node), isFirst(false){}
    };

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        stack<TagNode> stack;
        TreeNode* cur = root;
        while(cur != NULL || !stack.empty()){

            while(cur != NULL){
                stack.push(TagNode(cur));
                cur = cur->left;
            }

            TagNode tagNode = stack.top();
            stack.pop();
            cur = tagNode.node;
            if(tagNode.isFirst == false){
                tagNode.isFirst = true;
                stack.push(tagNode);
                cur = cur->right;
            }
            else{
                res.push_back(cur->val);
                cur = NULL;
            };
        }
        return res;
    }
};

int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-05-30

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// Non-Recursive
// Using two stacks, Reverse the Preorder Traversal!
//
// Time Complexity: O(n)
// Space Complexity: O(n)
class Solution {

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        stack<TreeNode*> stack, output;

        stack.push(root);
        while(!stack.empty()){

            TreeNode* node = stack.top();
            stack.pop();
            output.push(node);

            if(node->left != NULL)
                stack.push(node->left);
            if(node->right != NULL)
                stack.push(node->right);
        }

        while(!output.empty()){
            res.push_back((output.top())->val);
            output.pop();
        }

        return res;
    }
};

int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-05-30

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// Non-Recursive
// Using two stacks, Reverse the Preorder Traversal!
//
// Time Complexity: O(n)
// Space Complexity: O(n)
class Solution {

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        stack<TreeNode*> stack, output;

        TreeNode* p = root;
        while(p != NULL || !stack.empty()){
            if(p != NULL){
                stack.push(p);
                output.push(p);
                p = p->right;
            }
            else{
                p = stack.top();
                stack.pop();
                p = p->left;
            }
        }

        while(!output.empty()){
            res.push_back((output.top())->val);
            output.pop();
        }

        return res;
    }
};

int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-05-31

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// Non-Recursive
// Using a pre pointer to record the last visted node
//
// Time Complexity: O(n)
// Space Complexity: O(h)
class Solution {

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        stack<TreeNode*> stack;
        TreeNode* pre = NULL;

        stack.push(root);
        while(!stack.empty()){

            TreeNode* node = stack.top();
            stack.pop();
            if((node->left == NULL && node->right == NULL) ||
               (pre != NULL && pre == node->left && node->right == NULL) ||
               (pre != NULL && pre == node->right)){
                res.push_back(node->val);
                pre = node;
            }
            else{
                stack.push(node);

                if(node->right != NULL)
                    stack.push(node->right);
                if(node->left != NULL)
                    stack.push(node->left);
            }
        }

        return res;
    }
};


void print_vec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    TreeNode* root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->left = new TreeNode(3);
    print_vec(Solution().postorderTraversal(root));

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-05-31

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// Classic Non-Recursive
// Using a pre pointer to record the last visted node
//
// Time Complexity: O(n)
// Space Complexity: O(h)
class Solution {

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        stack<TreeNode*> stack;
        TreeNode* pre = NULL;
        TreeNode* cur = root;

        while(cur != NULL || !stack.empty()){

            while(cur != NULL){
                stack.push(cur);
                cur = cur->left;
            }

            cur = stack.top();
            stack.pop();

            if(cur->right == NULL || pre == cur->right){
                res.push_back(cur->val);
                pre = cur;
                cur = NULL;
            }
            else{
                stack.push(cur);
                cur = cur->right;
            }
        }

        return res;
    }
};


void print_vec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    TreeNode* root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->left = new TreeNode(3);
    print_vec(Solution().postorderTraversal(root));

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-05-31

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// Classic Non-Recursive
// Using a pre pointer to record the last visted node
//
// Time Complexity: O(n)
// Space Complexity: O(h)
class Solution {

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        stack<TreeNode*> stack;
        TreeNode* pre = NULL;
        TreeNode* cur = root;

        while(cur != NULL || !stack.empty()){
            if(cur != NULL){
                stack.push(cur);
                cur = cur->left;
            }
            else{
                cur = stack.top();
                stack.pop();

                if(cur->right == NULL || pre == cur->right){
                    res.push_back(cur->val);
                    pre = cur;
                    cur = NULL;
                }
                else{
                    stack.push(cur);
                    cur = cur->right;
                }
            }
        }

        return res;
    }
};


void print_vec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    TreeNode* root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->left = new TreeNode(3);
    print_vec(Solution().postorderTraversal(root));

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-05-31

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


// Morris PostOrder Traversal
//
// Time Complexity: O(n)
// Space Complexity: O(1)
class Solution {

public:
    vector<int> postorderTraversal(TreeNode* root) {

        vector<int> res;
        if(root == NULL)
            return res;

        TreeNode* dummyRoot = new TreeNode(-1);
        dummyRoot->left = root;

        TreeNode* cur = dummyRoot;
        while(cur != NULL){
            if(cur->left == NULL)
                cur = cur->right;
            else{
                TreeNode* prev = cur->left;
                while(prev->right != NULL && prev->right != cur)
                    prev = prev->right;

                if(prev->right == NULL){
                    prev->right = cur;
                    cur = cur->left;
                }
                else{
                    prev->right = NULL;
                    reverseTraversal(cur->left, res);
                    cur = cur->right;
                }
            }
        }
        delete dummyRoot;

        return res;
    }

private:
    void reverseTraversal(TreeNode* node, vector<int>& res){

        int start = res.size();
        while(node != NULL){
            res.push_back(node->val);
            node = node->right;
        }
        reverse(res.begin() + start, res.end());
    }
};


void print_vec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    TreeNode* root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->left = new TreeNode(3);
    print_vec(Solution().postorderTraversal(root));

    return 0;
}