LeetCode algorithm recursive class - Sword pointing to Offer 28. Symmetric binary tree

Table of contents

The sword points to Offer 28. Symmetric binary tree

answer:

Code:

operation result:

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The sword points to Offer 28. Symmetric binary tree

Please implement a function to determine whether a binary tree is symmetrical. A binary tree is symmetrical if it is the same as its mirror image.

For example, the 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 mirror symmetric:

    1
   / \
  2   2
   \   \
   3    3

Example 1:

Input: root = [1,2,2,3,4,4,3]
 Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
 Output: false

limit:

0 <= 节点个数 <= 1000

题解：

• First determine whether the root node is empty, and return true if it is empty.
• Then call the fun method and pass in the left child node and right child node of the root node for judgment.

The logic of the fun method is:

• If A and B are both empty, return true;
• If either A or B is empty or the values ​​are not equal, false is returned;
• Otherwise, recursively determine whether the left child node of A and the right child node of B are symmetrical, and whether the right child node of A and the left child node of B are symmetrical.
• Finally, the results are returned layer by layer through recursion, and a judgment result of whether it is symmetrical is obtained.

The time complexity is O(n)

代码：

/**
 * 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 fun(root.left,root.right);
    }
    public boolean fun(TreeNode A,TreeNode B){
        if(A==null&&B==null) return true;
        if(A==null||B==null||A.val!=B.val) return false;
        return fun(A.left,B.right)&&fun(A.right,B.left);
    }
}

运行结果：