101. Symmetric Tree

Description

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:

But the following [1,2,2,null,3,null,3] is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

Solution

Recursive

通过观察可知，对称树的根节点的左右子树是呈镜像的。

/**
 * 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 true;
        return isMirror(root.left, root.right);
    }
    
    public boolean isMirror(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;
        if (p == null || q == null) return false;
        return p.val == q.val && isMirror(p.left, q.right) && isMirror(p.right, q.left);
    }
}

Iterative

使用层序遍历的思路，对于每一层比较对称的节点。起初觉得需要两个队列，后来经提示发现一个队列就可以啦，在插入队列的顺序上做做文章，对称地插入队列就可以了。

/**
 * 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 true;
        Queue queue = new LinkedList();
        queue.add(root.left);
        queue.add(root.right);
        
        while (!queue.isEmpty()) {
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();
            if (left == null && right == null) continue;
            if (left == null || right == null || left.val != right.val) return false;
            queue.add(left.left);
            queue.add(right.right);
            queue.add(left.right);
            queue.add(right.left);
        }
        
        return true;
    }
}