Showing posts with label inorder. Show all posts
Showing posts with label inorder. Show all posts

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

[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
        (c) Push the top node's right child into stack
        (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.
        (c) Push the top node's left child into stack 1.
        (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 75: Recover Binary Search Tree

Recover Binary Search Tree


Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

Analysis:


The question requires O(1) space, let's first consider the case without this.
How to traverse BST?
Yep! The inOrder tree traversal.  (left->root->right)
So the easiest way is inorder traverse the BST and find the element pair (two elements) which are not consistent with the definition of BST. In order to get the order, a queue is needed, which is O(n).

Now how to do this procedure in O(1)?
What we need is actually two pointers, which point to 2 tree nodes where is incorrect. Therefore, we only need to store these two pointers, and, we also need another pointer to store the previous element, in order to  compare if the current element is valid or not.

In the code below, you will find, the last step is to replace the wrong pair's value. And the inOrder function is to search the whole BST and find the wrong pairs.

Note that, (1)the previous element is NOT the root node of the current element, but the previous element in the "inOrder" order; (2) To store the wrong pair, the first found wrong element is stored in first pointer, while the next is stored in the second pointer.

e.g. The correct BST is below:

The inorder traversal is :  1 3 4 6 7 8 10 13 14

If we change the value 4 and 8:  1 3 8 6 7 4 10 13 14, when it goes to the node 6, now the pre->val = 8, check if pre<current node, which is wrong here (8>6). So we store the first pointer pointing to the node 6 and second pointer pointing to the node 8. To do so, we have stored the wrong nodes, if every other node keep the correct order, then swapping these nodes is enough for the problem. In other words, after the whole traversal, what we need to do is just changing the values of the first and second. Continue our example here, when the traversal goes to the node 4, now the node 7 is its pre, which is also wrong, so the second wrong node is found, and we change the second pointer pointing to node 4.



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:
    TreeNode *first;
    TreeNode *second;
    TreeNode *pre;
    
    void inOrder(TreeNode *root){
        if (root==NULL){return;}
        else{
            inOrder(root->left);
            if (pre == NULL){pre = root;}
            else {
                if (pre->val > root->val){
                    if (first==NULL) {first = pre;}
                    second = root;
                }
                pre = root;
            }
            inOrder(root->right);
            
        }
    }
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        pre = NULL;
        first = NULL;
        second= NULL;
        inOrder(root);
        int val;
        val = first->val;
        first->val=second->val;
        second->val=val;
        return;
    }
};

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 tree node
    prev = None
    p1 = None
    p2 = None
    
    def inOrder(self, root):
        if root == None:
            return
        else:
            self.inOrder(root.left)
            if self.prev == None:
                self.prev = root
            else:
                if root.val <= self.prev.val:
                    if self.p1 == None:
                        self.p1 = self.prev
                    self.p2 = root
                self.prev = root
            self.inOrder(root.right)
    
    def recoverTree(self, root):
        self.prev = None
        self.p1 = None
        self.p2 = None
        self.inOrder(root)
        tmp = self.p1.val
        self.p1.val = self.p2.val
        self.p2.val = tmp
        return root