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101.Symmetric-Tree

101. Symmetric Tree

题目地址

题目描述

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.

代码

Approach #1 Recursive

Time complexity & Space complexity: O(n)
/**
* 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) {
return isMirror(root, root);
}
​
public boolean isMirror(TreeNode t1, TreeNode t2) {
if (t1 == null && t2 == null) return true;
if (t1 == null || t2 == null) return false;
return (t1.val == t2.val)
&& isMirror(t1.right, t2.left)
&& isMirror(t1.left, t2.right);
}
}

Approach #2 Iterative

class Solution {
public boolean isSymetric(TreeNode root) {
Queue<TreeNode> queue = new LinkedList();
queue.add(root);
queue.add(root);
​
while (!queue.isEmpty()) {
TreeNode t1 = queue.poll();
TreeNode t2 = queue.poll();
if (t1 == null && t2 == null) continue;
if (t1 == null || t2 == null) return false;
if (t1.val != t2.val) return false;
queue.add(t1.left);
queue.add(t2.right);
queue.add(t1.right);
queue.add(t2.left);
}
​
return true;
}
}