Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3]
is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3]
is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
if(root!=null){
return isMirror(root.left,root.right);
}
return true;
}
public boolean isMirror(TreeNode r1,TreeNode r2){
if(r1==null&&r2==null){
return true;
}
if(r1==null||r2==null){
return false;
}
if(r1.val==r2.val){
return isMirror(r1.left,r2.right)&&isMirror(r1.right,r2.left);
}
return false;
}
}
Determine whether a number is symmetric, we need to compare two nodes, and the declaration of a isMirror method to determine the left and right node are empty, or about a node is empty, we only excluded r1, r2 is empty case before they can be r1, r2 comparison value, or a null pointer exception will be reported. When a node is about equal to the value of time, we continue to the bottom of the recursive call isMirror method to arrive at the answer