Given two binary trees, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
Example 1:
Input: 1 1
/ \ / \
2 3 2 3
[1,2,3], [1,2,3]
Output: true
Example 2:
Input: 1 1
/ \
2 2
[1,2], [1,null,2]
Output: false
Example 3:
Input: 1 1
/ \ / \
2 1 1 2
[1,2,1], [1,1,2]
Output: false
- Iterative Approach
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} p
* @param {TreeNode} q
* @return {boolean}
*/
var isSameTree = function(p, q) {
let stack1 = [p]
let stack2 = [q]
// Base Case
if (p == null && q == null) return true
if (p == null && q !== null) return false
if (p !== null && q == null) return false
//Tree Traversal for p & q simultaneously
while (stack1.length) {
//Unload Stack
let tn1 = stack1.pop()
let tn2 = stack2.pop()
//Check if node have same value
if (tn1.val !== tn2.val) return false
//Check if node have same structure - null hell
if (tn1.left == null && tn2.left != null) {
//left
return false
} else if (tn1.left != null && tn2.left == null) {
//left
return false
} else if (tn1.right == null && tn2.right != null) {
//right
return false
} else if (tn1.right != null && tn2.right == null) {
//right
return false
}
//Assuming all same, traverse tree using stack
if (tn1.right) stack1.push(tn1.right)
if (tn2.right) stack2.push(tn2.right)
if (tn1.left) stack1.push(tn1.left)
if (tn2.left) stack2.push(tn2.left)
}
return true
}
- Time Complexity:
O(n)
- space Complexity:
O(n)