对称二叉树

题目

给定一个二叉树，检查它是不是镜像对称的。

例如，二叉树[1,2,2,3,4,4,3]是对称的。

1
   / \
  2   2
 / \ / \
3  4 4  3

可是下面这个[1,2,2,null,3,null,3]则不是镜像对称的:

1
   / \
  2   2
   \   \
   3    3

解法

解法一： 递归。

/** * 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 { // 方法一：递归 public boolean isSymmetric(TreeNode root) { if(root == null){ return true; } TreeNode left = root.left; TreeNode right = root.right; return isSymmetricForLeftAndRight(left,right); } // 判断left和right是否对称 public boolean isSymmetricForLeftAndRight(TreeNode left,TreeNode right){ if(left == null || right == null){ if(left != null || right != null){ return false; } else { return true; } } // 判断值是否相等 int leftVal = left.val; int rightVal = right.val; if(leftVal != rightVal){ return false; } // 递归调用，判断左节点的左节点与右节点的右节点是否对称 TreeNode leftNextLeft = left.left; TreeNode rightNextRight = right.right; if(!isSymmetricForLeftAndRight(leftNextLeft,rightNextRight)){ return false; } // 递归调用，判断左节点的右节点与右节点的左节点是否对称 TreeNode leftNextRight = left.right; TreeNode rightNextLeft = right.left; if(!isSymmetricForLeftAndRight(leftNextRight,rightNextLeft)){ return false; } return true; } } 

解法二： 迭代。

/** * 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 { // 解法二：迭代 public boolean isSymmetric(TreeNode root) { if(root == null){ return true; } // 建立队列，特色是先进先出。 Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root.left); queue.offer(root.right); while(!queue.isEmpty()){ TreeNode left = queue.poll(); TreeNode right = queue.poll(); if(left == null || right == null){ if(left != null || right != null){ return false; } else { continue; } } int leftVal = left.val; int rightVal = right.val; if(leftVal != rightVal){ return false; } // 再把left和right子节点放进去，注意顺序 // 顺序必定是left的左节点和right的右节点，而后是left的右节点和right的左节点 queue.offer(left.left); queue.offer(right.right); queue.offer(left.right); queue.offer(right.left); } return true; } } 

总结

本篇文章讲解了算法题目的思路和解法，代码和笔记因为纯手打，不免会有纰漏，若是发现错误的地方，请第一时间告诉我，这将是我进步的一个很重要的环节。之后会按期更新算法题目以及各类开发知识点，若是您以为写得不错，不妨点个关注，谢谢。