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[Re-view] Tree Traversal (PreOrder, InOrder, PostOrder): Recursion and Non-Recursion(Iteration)

Introduction

In this post, I would like to talk one important algorithm---- Tree Traversal. Tree traversal is a basic but crucial concept in data structure and algorithm, which is also very hot in interviews. In this post, I assume that you have already had the idea of "what is binary tree?", "what is tree traversal?", and "what is stack?". I will focus on Recursive and Non-Recursive traversal, for PreOrder, InOrder, PostOrder, and Level Order of binary tree. Algorithm description with GIF animation are provided along with the C++ implementations. The code can be run and tested in leetcode online judge as well.

Actually the "three order" traversal is easily to memorize in this way: Where is the root? When the root is in the first place to be visited (root->left->right), then it is "pre" order, when the root is in the middle to be visited (left->root->right), then it is "in" order, and when the root is last to be visited, it is "post" order.

PreOrder Traversal:

Description:

PreOrder traversal is according to the root, left and right order traversing the binary tree.

Recursive traversal is straightforward: we first visit the root node, then traverse its left child, then its right child.

Non-Recursive traversal is to ulitize the data structure stack. There are three steps for this algorithm:

(1) Initialize the stack. Push the root node into stack.

(2) While the stack is not empty do:

(a) Pop the top node from the stack.

(b) Store the top node

(d) Push the top node's left child into stack

(3) Output the stored sequence.

Animation(non-recursive):

Code (PreOrder Traversal):

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
      // preorder traversal: recursion 
    void pre_recur(TreeNode* root, vector<int> &res){
        if (!root){         //if null node, return
            return;
        }else{
            res.push_back(root->val); // traverse root child
            if (root->left){    
                pre_recur(root->left,res); //traverse left child
            }
            if (root->right){
                pre_recur(root->right,res); //traverse right child
            }
        }
    }
    
    // preorder traversal: non-recursion
    void pre_norecur(TreeNode* root,vector<int> &res){
        stack<TreeNode*> st;
        st.push(root);   //initialize stack
        
        TreeNode* tmp;
        while (!st.empty()){
            tmp=st.top();  //get the top node in stack 
            st.pop();
            res.push_back(tmp->val); //store the root value
            if (tmp->right){st.push(tmp->right);} //push right child into stack 
            if (tmp->left){st.push(tmp->left);} //push left child into stack 
        }
    }
    
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> res;
        if (!root){return res;}
        
        //pre_recur(root,res);   //recursive traversal 
        pre_norecur(root,res); //non-recursive traversal
        return res;
    }
};

InOrder Traversal:

Description:

InOrder traversal is according to the left, root, and right order traversing the binary tree.

Recursive traversal is straightforward: we first traverse the left child, then visit the root node, then traverse the right child.

Non-Recursive traversal is to ulitize the data structure stack. There are three steps for this algorithm:

(1) Initialize the stack. Set root as the current node. Push the current node into stack.

(2) Loop until finished (while true):

(a) If current has left child:

(i)Push left child into stack.

(ii)Current node = current node's left child

(b) If current has NO left child:

(i)If stack is empty: exit

(ii)If stack is NOT empty:

Pop the top node in the stack.

Store the top node for output.

Set current node = top node's right child.

(3) Output the stored sequence.

Animation(non-recursive):

Code (InOrder Traversal):

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    // inorder traversal: recursion 
    void in_recur(TreeNode* root, vector<int> &res){
        if (!root){         //if null node, return
            return;
        }else{
            if (root->left){in_recur(root->left,res);} //traverse left child
            res.push_back(root->val); // traverse root child
            if (root->right){in_recur(root->right,res);} //traverse right child
        }
    }
    
    // inorder traversal: non-recursion
    void in_norecur(TreeNode* root,vector<int> &res){
        stack<TreeNode*> st;
        while (true){
            if (root){  // keep pushing the left child of current node
                st.push(root);
                root=root->left;
            }else{
                if (!st.empty()){   
                    TreeNode* top = st.top();   //pop the top node in the stack
                    st.pop();
                    res.push_back(top->val);    //output the top node
                    root=top->right;            //set current node=right child of top node
                }else{
                    break;  //if the stack is empty, then exit
                }
            }
        }
    }

    vector<int> inorderTraversal(TreeNode *root) {
       vector<int> res;
        if (!root){return res;}
        //in_recur(root,res);   //recursive traversal 
        in_norecur(root,res); //non-recursive traversal
        return res;
    }
};

PostOrder Traversal:

Description:

PostOrder traversal is according to the left, right, and root order traversing the binary tree.

Recursive traversal is straightforward: we first traverse the left child, then traverse the right child, then visit the root node.

Non-Recursive traversal is to ulitize the data structure stack. Two stacks is used in this algorithm. Stack 2 is used to store the result. There are three steps:

(1) Initialize the stack. Push the root node into stack.

(2) While the stack is not empty do:

(a) Pop the top node from the stack 1.

(b) Push the top node into stack 2.

(d) Push the top node's right child into stack 1.

(3) Pop all the nodes in Stack 2 to get the result.

Animation(non-recursive):

Code (PostOrder Traversal):

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    // postorder traversal: recursion 
    void post_recur(TreeNode* root, vector<int> &res){
        if (!root){         //if null node, return
            return;
        }else{
            if (root->left){    
                post_recur(root->left,res); //traverse left child
            }
            if (root->right){
                post_recur(root->right,res); //traverse right child
            }
            res.push_back(root->val); // traverse root child
        }
    }
    
    // postorder traversal: non-recursion
    void post_norecur(TreeNode* root,vector<int> &res){
        stack<TreeNode*> st1;
        stack<TreeNode*> st2;
        st1.push(root);   //initialize stack one
        
        TreeNode* tmp;
        while (!st1.empty()){
            tmp=st1.top();  //get the top node in stack 1
            st1.pop();  
            st2.push(tmp);  //push into stack 2
            if (tmp->left){st1.push(tmp->left);} //push left child into stack 1
            if (tmp->right){st1.push(tmp->right);} //push right child into stack 1
        }
        
        while (!st2.empty()){  //output the stack 2
            tmp=st2.top();
            st2.pop();
            res.push_back(tmp->val);
        }
        
    }

    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> res;
        if (!root){return res;}
        
        //post_recur(root,res);   //recursive traversal 
        post_norecur(root,res); //non-recursive traversal
        return res;
    }
};

leetcode Question: Binary Tree Postorder Traversal (iteration)

Binary Tree Postorder Traversal (iteration)

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

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

return [3,2,1].

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

Analysis:

Classical problem, the recursion version of the solution can be easily found by modifying here. In this post, I solved it iteratively. Which data structure do remind us ? Yes! Still stack!

The algorithm has following steps, which is slight different from the previous question:
(1) Push the root node into the stack.
(2) while the stack is not empty, do:
if
the top node is a leaf node (no left&right children), pop it.
or if the top node has been visited, pop it. (here we use a sign head to show the latest popped node, if the top node's child = the latest popped node, either left or right, it must has been visited.)
else
b. push the right child of the top node (if exists)
c. push the left child of the top node (if exists)

Code (C++):

/**
 * Definition for binary tree
 * 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) {
        stack<TreeNode*> st;
        vector<int> res;
        if (!root){return res;}
        st.push(root);
        TreeNode* head=root;
        while (!st.empty()){
            TreeNode* top = st.top();
            if ((top->left==NULL && top->right==NULL)||top->right==head || top->left==head){
                res.push_back(top->val);
                st.pop();
                head = top;
            }else{
                if (top->right!=NULL){st.push(top->right);}
                if (top->left!=NULL){st.push(top->left);}
            }
        }
        return res;
    }
};

Code(Python):

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def postorderTraversal(self, root):
        res = []
        st = []
        if root != None:
            st.append([root,False])
            while len(st) != 0:
                tmp = st[-1]
                if tmp[1] == True:
                    res.append(st.pop()[0].val)
                elif tmp[0].left == None and tmp[0].right == None:
                    res.append(st.pop()[0].val)
                else:
                    st[-1][1] = True
                    if tmp[0].right != None:
                        st.append([tmp[0].right, False])
                    if tmp[0].left != None:
                        st.append([tmp[0].left, False])
        return res

leetcode Question: Binary Tree Preorder Traversal (iteration)

Binary Tree Preorder Traversal (iteration)

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

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

return [1,2,3].

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

Analysis:

Classical problem, the recursion version of the solution can be found here (you need do slight modification, can you do that?) . In this post, I solved it iteratively. Which data structure do remind us ? Yes! Stack!

The algorithm has following steps:
(1) Push the root node into the stack.
(2) while the stack is not empty, do:
a. pop the top node and print it.
b. push the right child of the top node (if exists)
c. push the left child of the top node (if exists)

Code(C++):

/**
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        stack<TreeNode*> st;
        vector<int> res;
        if (!root){return res;}
        st.push(root);
        TreeNode* head=root;
        while (!st.empty()){
            TreeNode* top = st.top();
            res.push_back(top->val);
            st.pop();
            if (top->right!=NULL){st.push(top->right);}
            if (top->left!=NULL){st.push(top->left);}
        }
        return res;
    }
};

Code(Python):

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def preorderTraversal(self, root):
        res = [] 
        st = []
        if root != None:
            st.append(root)
            while len(st) != 0:
                tmp = st.pop()
                res.append(tmp.val)
                if (tmp.right != None):
                    st.append(tmp.right)
                if (tmp.left != None):
                    st.append(tmp.left)
        return res