https://leetcode.cn/problems/symmetric-tree/
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3]
Output: false
Constraints:
The number of nodes in the tree is in the range [1, 1000].
-100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root){
Queue<TreeNode> q = new LinkedList<TreeNode>();
TreeNode L = root.left;
TreeNode R = root.right;
if(root == null || L == null && R == null){
return true;
}
q.add(L);
q.add(R);
while(!q.isEmpty()){
L = q.remove();
R = q.remove();
if(L == null && R == null){
continue;
}
if( (L == null || R == null) || (L.val != R.val) ){
return false;
}
q.add(L.left);
q.add(R.right);
q.add(L.right);
q.add(R.left);
}
return true;
}
// public boolean isSymmetric(TreeNode root) {
// if(root == null){
// return true;
// }
// //invoke the recursive function to compare the left subtree and the right subtree
// return deepCheck(root.left, root.right);
// }
private boolean deepCheck(TreeNode left, TreeNode right){
//Conditions to exit: both are null
//or either is null
//or vavlues of two node are unequal
if(left == null && right == null){
return true;
}
if(left == null || right == null){
return false;
}
if(left.val != right.val){
return false;
}
//compare recursivly, left child of left vs right child of right,
//and right child of left vs left child of right
return deepCheck(left.left, right.right) && deepCheck(left.right, right.left);
}
}