• Symmetric Tree(对称二叉树)


    来源:https://leetcode.com/problems/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

    递归:

    1. 若根节点为null,返回true

    2. 若根节点不为null,对其左右子树进行递归地比较:若两个节点都为null,返回true;若一个为null,另一个不为null,返回false;若两个节点(A,B)值相等,则返回 (A.left, B.right) 的比较结果 && (A.right, B.left) 的比较结果。

    Java

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 class Solution {
    11     public boolean compareLR(TreeNode left, TreeNode right) {
    12         if(left == null && right == null) {
    13             return true;
    14         }
    15         if((left == null && right != null) || (left != null && right == null)) {
    16             return false;
    17         }
    18         if(left.val == right.val) {
    19             return compareLR(left.right, right.left) && compareLR(left.left, right.right);
    20         }
    21         return false;
    22     }
    23     public boolean isSymmetric(TreeNode root) {
    24         if(root == null) {
    25             return true;
    26         }
    27         return compareLR(root.left, root.right);
    28     }
    29 }

    Python

     1 # -*- coding:utf-8 -*-
     2 # class TreeNode:
     3 #     def __init__(self, x):
     4 #         self.val = x
     5 #         self.left = None
     6 #         self.right = None
     7 class Solution:
     8     def isSymmetrical(self, pRoot):
     9         def comRoot(left, right):
    10             if not left:
    11                 return not right
    12             if not right or (left.val != right.val):
    13                 return False
    14             return comRoot(left.right, right.left) and comRoot(left.left, right.right)
    15         if not pRoot:
    16             return True
    17         return comRoot(pRoot.left, pRoot.right)

    迭代:

    1. 若根节点为null,返回true

    2. 若根节点不为null,将其左右两个子节点入队列

    3. 从队列中取出两个节点,若两个节点都为null,返回true;若一个为null,另一个不为null,返回false;若两个节点(A,B)值相等,将他们的两个子节点按照A.left,B.right,A.right,B.left的顺序入队列,重复执行该步,直到队列中不再有节点。

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  • 原文地址:https://www.cnblogs.com/renzongxian/p/7517677.html
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