Problem
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
Solution 1
/**
* 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) {
if ((!p && q) || (p && !q) || (p && q && p.val !== q.val)) return false;
if (p && q) return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
return true;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
n为节点数。 - Space complexity : O(1).
Solution 2
/**
* 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) {
var s1 = [p];
var s2 = [q];
var ll = null;
var rr = null;
while (s1.length && s2.length) {
ll = s1.pop();
rr = s2.pop();
if (!ll && !rr) continue;
if (!ll || !rr) return false;
if (ll.val !== rr.val) return false;
s1.push(ll.left);
s1.push(ll.right);
s2.push(rr.left);
s2.push(rr.right);
}
return true;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
n为节点数。 - Space complexity : O(n).