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  • [LC] 951. Flip Equivalent Binary Trees

    For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

    A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

    Write a function that determines whether two binary trees are flip equivalent.  The trees are given by root nodes root1 and root2.

    Example 1:

    Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
    Output: true
    Explanation: We flipped at nodes with values 1, 3, and 5.
    Note:
    1. Each tree will have at most 100 nodes.
    2. Each value in each tree will be a unique integer in the range [0, 99].

    Time: O(N)

    /**
     * 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 {
        public boolean flipEquiv(TreeNode root1, TreeNode root2) {
            if (root1 == null || root2 == null) {
                return root1 == root2;
            }
            if (root1.val != root2.val) {
                return false;
            }
            return flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right) 
                || flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left);
    
        }
    }
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  • 原文地址:https://www.cnblogs.com/xuanlu/p/13063083.html
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