Tree Traversals (Inorder, Preorder and Postorder)
Unlike linear data structures (Array, Linked List, Queues, Stacks, etc) which have only one logical way to traverse them, trees can be traversed in different ways. The following are the generally used methods for traversing trees:
Example:
Inorder Traversal (Practice):
Algorithm Inorder(tree)
- Traverse the left subtree, i.e., call Inorder(left->subtree)
- Visit the root.
- Traverse the right subtree, i.e., call Inorder(right->subtree)
Uses of Inorder Traversal:
In the case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. To get nodes of BST in non-increasing order, a variation of Inorder traversal where Inorder traversal is reversed can be used.
Example: In order traversal for the above-given figure is 4 2 5 1 3.
C++
// C++ program for different tree traversals#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left childand a pointer to right child */struct Node { int data; struct Node *left, *right;};// Utility function to create a new tree nodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}/* Given a binary tree, print its nodes in inorder*/void printInorder(struct Node* node){ if (node == NULL) return; /* first recur on left child */ printInorder(node->left); /* then print the data of node */ cout << node->data << " "; /* now recur on right child */ printInorder(node->right);}/* Driver code*/int main(){ struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); // Function call cout << "\nInorder traversal of binary tree is \n"; printInorder(root); return 0;} |
C
// C program for different tree traversals#include <stdio.h>#include <stdlib.h>/* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right;};/* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data){ struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node);}/* Given a binary tree, print its nodes in inorder*/void printInorder(struct node* node){ if (node == NULL) return; /* first recur on left child */ printInorder(node->left); /* then print the data of node */ printf("%d ", node->data); /* now recur on right child */ printInorder(node->right);}/* Driver code*/int main(){ struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); // Function call printf("\nInorder traversal of binary tree is \n"); printInorder(root); getchar(); return 0;} |
Java
// Java program for different tree traversals/* Class containing left and right child of current node and key value*/class Node { int key; Node left, right; public Node(int item) { key = item; left = right = null; }}class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in inorder*/ void printInorder(Node node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ System.out.print(node.key + " "); /* now recur on right child */ printInorder(node.right); } // Wrappers over above recursive functions void printInorder() { printInorder(root); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); // Function call System.out.println( "\nInorder traversal of binary tree is "); tree.printInorder(); }} |
Python
# Python3 program to for tree traversals# A class that represents an individual node in a# Binary Treeclass Node: def __init__(self, key): self.left = None self.right = None self.val = key# A function to do inorder tree traversaldef printInorder(root): if root: # First recur on left child printInorder(root.left) # then print the data of node print(root.val), # now recur on right child printInorder(root.right)# Driver codeif __name__ == "__main__": root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) # Function call print "\nInorder traversal of binary tree is" printInorder(root) |
C#
// C# program for different// tree traversalsusing System;/* Class containing left andright child of currentnode and key value*/class Node { public int key; public Node left, right; public Node(int item) { key = item; left = right = null; }}class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in inorder*/ void printInorder(Node node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ Console.Write(node.key + " "); /* now recur on right child */ printInorder(node.right); } // Wrappers over above recursive functions void printInorder() { printInorder(root); } // Driver Code static public void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); // Function call Console.WriteLine("\nInorder traversal " + "of binary tree is "); tree.printInorder(); }}// This code is contributed by Arnab Kundu |
Javascript
<script>// javascript program for different tree traversals/* Class containing left and right child of current node and key value*/class Node { constructor(val) { this.key = val; this.left = null; this.right = null; }} // Root of Binary Tree var root = null; /* * Given a binary tree, print its nodes according to the "bottom-up" postorder * traversal. */ function printPostorder(node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node document.write(node.key + " "); } /* Given a binary tree, print its nodes in inorder */ function printInorder(node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ document.write(node.key + " "); /* now recur on right child */ printInorder(node.right); } /* Given a binary tree, print its nodes in preorder */ function printPreorder(node) { if (node == null) return; /* first print data of node */ document.write(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Driver method root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); document.write("Preorder traversal of binary tree is <br/>"); printPreorder(root); document.write("<br/>Inorder traversal of binary tree is <br/>"); printInorder(root); document.write("<br/>Postorder traversal of binary tree is <br/>"); printPostorder(root);// This code is contributed by aashish1995</script> |
Output
Inorder traversal of binary tree is 4 2 5 1 3
Time Complexity: O(N)
Auxiliary Space: If we don’t consider the size of the stack for function calls then O(1) otherwise O(h) where h is the height of the tree.
Preorder Traversal (Practice):
Algorithm Preorder(tree)
- Visit the root.
- Traverse the left subtree, i.e., call Preorder(left->subtree)
- Traverse the right subtree, i.e., call Preorder(right->subtree)
Uses of Preorder:
Preorder traversal is used to create a copy of the tree. Preorder traversal is also used to get prefix expressions on an expression tree.
Example: Preorder traversal for the above-given figure is 1 2 4 5 3.
C++
// C++ program for different tree traversals#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left childand a pointer to right child */struct Node { int data; struct Node *left, *right;};// Utility function to create a new tree nodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}/* Given a binary tree, print its nodes in preorder*/void printPreorder(struct Node* node){ if (node == NULL) return; /* first print data of node */ cout << node->data << " "; /* then recur on left subtree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right);}/* Driver code*/int main(){ struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); // Function call cout << "\nPreorder traversal of binary tree is \n"; printPreorder(root); return 0;} |
C
// C program for different tree traversals#include <stdio.h>#include <stdlib.h>/* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right;};/* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data){ struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node);}/* Given a binary tree, print its nodes in preorder*/void printPreorder(struct node* node){ if (node == NULL) return; /* first print data of node */ printf("%d ", node->data); /* then recur on left subtree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right);}/* Driver code*/int main(){ struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); // Function call printf("\nPreorder traversal of binary tree is \n"); printPreorder(root); getchar(); return 0;} |
Java
// Java program for different tree traversals/* Class containing left and right child of current node and key value*/class Node { int key; Node left, right; public Node(int item) { key = item; left = right = null; }}class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in preorder*/ void printPreorder(Node node) { if (node == null) return; /* first print data of node */ System.out.print(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPreorder() { printPreorder(root); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); // Function call System.out.println( "Preorder traversal of binary tree is "); tree.printPreorder(); }} |
Python
# Python3 program to for tree traversals# A class that represents an individual node in a# Binary Treeclass Node: def __init__(self, key): self.left = None self.right = None self.val = key# A function to do preorder tree traversaldef printPreorder(root): if root: # First print the data of node print(root.val), # Then recur on left child printPreorder(root.left) # Finally recur on right child printPreorder(root.right)# Driver codeif __name__ == "__main__": root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) # Function call print "Preorder traversal of binary tree is" printPreorder(root) |
C#
// C# program for different// tree traversalsusing System;/* Class containing left andright child of currentnode and key value*/class Node { public int key; public Node left, right; public Node(int item) { key = item; left = right = null; }}class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in preorder*/ void printPreorder(Node node) { if (node == null) return; /* first print data of node */ Console.Write(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPreorder() { printPreorder(root); } // Driver Code static public void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); // Function call Console.WriteLine("Preorder traversal " + "of binary tree is "); tree.printPreorder(); }}// This code is contributed by Arnab Kundu |
Javascript
<script>// javascript program for different tree traversals/* Class containing left and right child of current node and key value*/class Node { constructor(val) { this.key = val; this.left = null; this.right = null; }} // Root of Binary Tree var root = null; /* * Given a binary tree, print its nodes according to the "bottom-up" postorder * traversal. */ function printPostorder(node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node document.write(node.key + " "); } /* Given a binary tree, print its nodes in inorder */ function printInorder(node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ document.write(node.key + " "); /* now recur on right child */ printInorder(node.right); } /* Given a binary tree, print its nodes in preorder */ function printPreorder(node) { if (node == null) return; /* first print data of node */ document.write(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Driver method root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); document.write("Preorder traversal of binary tree is <br/>"); printPreorder(root); document.write("<br/>Inorder traversal of binary tree is <br/>"); printInorder(root); document.write("<br/>Postorder traversal of binary tree is <br/>"); printPostorder(root);// This code is contributed by aashish1995</script> |
Output
Preorder traversal of binary tree is 1 2 4 5 3
Time Complexity: O(N)
Auxiliary Space: If we don’t consider the size of the stack for function calls then O(1) otherwise O(h) where h is the height of the tree.
Postorder Traversal (Practice):
Algorithm Postorder(tree)
- Traverse the left subtree, i.e., call Postorder(left->subtree)
- Traverse the right subtree, i.e., call Postorder(right->subtree)
- Visit the root
Uses of Postorder:
Postorder traversal is used to delete the tree. Please see the question for the deletion of a tree for details. Postorder traversal is also useful to get the postfix expression of an expression tree
Example: Postorder traversal for the above-given figure is 4 5 2 3 1
Below is the implementation of the above traversal methods:
C++
// C++ program for different tree traversals#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left childand a pointer to right child */struct Node { int data; struct Node *left, *right;};// Utility function to create a new tree nodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}/* Given a binary tree, print its nodes according to the"bottom-up" postorder traversal. */void printPostorder(struct Node* node){ if (node == NULL) return; // first recur on left subtree printPostorder(node->left); // then recur on right subtree printPostorder(node->right); // now deal with the node cout << node->data << " ";}/* Driver code*/int main(){ struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); // Function call cout << "\nPostorder traversal of binary tree is \n"; printPostorder(root); return 0;} |
C
// C program for different tree traversals#include <stdio.h>#include <stdlib.h>/* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right;};/* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data){ struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node);}/* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */void printPostorder(struct node* node){ if (node == NULL) return; // first recur on left subtree printPostorder(node->left); // then recur on right subtree printPostorder(node->right); // now deal with the node printf("%d ", node->data);}/* Driver code*/int main(){ struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); // Function call printf("\nPostorder traversal of binary tree is \n"); printPostorder(root); getchar(); return 0;} |
Java
// Java program for different tree traversals/* Class containing left and right child of current node and key value*/class Node { int key; Node left, right; public Node(int item) { key = item; left = right = null; }}class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */ void printPostorder(Node node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node System.out.print(node.key + " "); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); // Function call System.out.println( "\nPostorder traversal of binary tree is "); tree.printPostorder(); }} |
Python
# Python3 program to for tree traversals# A class that represents an individual node in a# Binary Treeclass Node: def __init__(self, key): self.left = None self.right = None self.val = key# A function to do postorder tree traversaldef printPostorder(root): if root: # First recur on left child printPostorder(root.left) # the recur on right child printPostorder(root.right) # now print the data of node print(root.val),# Driver codeif __name__ == "__main__": root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) # Function call print "\nPostorder traversal of binary tree is" printPostorder(root) |
C#
// C# program for different// tree traversalsusing System;/* Class containing left andright child of currentnode and key value*/class Node { public int key; public Node left, right; public Node(int item) { key = item; left = right = null; }}class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */ void printPostorder(Node node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node Console.Write(node.key + " "); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } // Driver Code static public void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); // Function call Console.WriteLine("\nPostorder traversal " + "of binary tree is "); tree.printPostorder(); }}// This code is contributed by Arnab Kundu |
Javascript
<script>// javascript program for different tree traversals/* Class containing left and right child of current node and key value*/class Node { constructor(val) { this.key = val; this.left = null; this.right = null; }} // Root of Binary Tree var root = null; /* * Given a binary tree, print its nodes according to the "bottom-up" postorder * traversal. */ function printPostorder(node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node document.write(node.key + " "); } /* Given a binary tree, print its nodes in inorder */ function printInorder(node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ document.write(node.key + " "); /* now recur on right child */ printInorder(node.right); } /* Given a binary tree, print its nodes in preorder */ function printPreorder(node) { if (node == null) return; /* first print data of node */ document.write(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Driver method root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); document.write("Preorder traversal of binary tree is <br/>"); printPreorder(root); document.write("<br/>Inorder traversal of binary tree is <br/>"); printInorder(root); document.write("<br/>Postorder traversal of binary tree is <br/>"); printPostorder(root);// This code is contributed by aashish1995</script> |
Output
Postorder traversal of binary tree is 4 5 2 3 1
Time Complexity: O(N)
Auxiliary Space: If we don’t consider the size of the stack for function calls then O(1) otherwise O(h) where h is the height of the tree.
Note: The height of the skewed tree is n (no. of elements) so the worst space complexity is O(N) and the height is (Log N) for the balanced tree so the best space
complexity is O(Log N).
Let us see different corner cases:
Complexity function T(n) — for all problems where tree traversal is involved — can be defined as: T(n) = T(k) + T(n – k – 1) + c
Where k is the number of nodes on one side of the root and n-k-1 on the other side.
Let’s do an analysis of boundary conditions:
Case 1: Skewed tree (One of the subtrees is empty and another subtree is non-empty )
k is 0 in this case.
T(n) = T(0) + T(n-1) + c
T(n) = 2T(0) + T(n-2) + 2c
T(n) = 3T(0) + T(n-3) + 3c
T(n) = 4T(0) + T(n-4) + 4c
…………………………………………
………………………………………….
T(n) = (n-1)T(0) + T(1) + (n-1)c
T(n) = nT(0) + (n)c
Value of T(0) will be some constant say d. (traversing an empty tree will take some constants time)
T(n) = n(c+d)
T(n) = Θ(n) (Theta of n)Case 2: Both left and right subtrees have an equal number of nodes.
T(n) = 2T(|_n/2_|) + c
This recursive function is in the standard form (T(n) = aT(n/b) + (-)(n) ) for master method. If we solve it by master method we get (-)(n)
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