Leetcode-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:

    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.

分析

如果一个树是镜像树，那么这颗树有如下特点，

1. 左子树的 left-root-right 遍历和右子树的 right-root-left 遍历结果一致
2. 左子树的左节点和右子树的右节点值相等，并且左子树的左子树是右子树的右子树的镜像
3. 左子树的右节点和右子树的左节点值相等，并且左子树的右子树是右子树的左子树的镜像

根据第 1 条特点可以写出非递归算法，根据 2、3 两条可以写出递归算法。

递归写法

根据分析得出的 2、3 特性，代码实现如下；

/**
 * 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 root == null || isIndentical(root.left, root.right);
    }
    
    private boolean isIndentical(TreeNode front, TreeNode back) {
        if (front == null || back == null) {
            return front == back;
        }
        if (back.val != front.val) {
            return false;
        }
        
        return isIndentical(front.right, back.left) 
            && isIndentical(front.left, back.right);
    }
}

提交后，得分不高，运行时打败 31%。

非递归解法

未完，待续…