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113th LeetCode Weekly Contest Flip Equivalent Binary Trees

时间:2018-12-02 22:51:17      阅读:243      评论:0      收藏:0      [点我收藏+]

标签:tree node   左右子树   ati   unique   opera   ons   make   output   def   

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.
技术分享图片

 

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].

A、B两颗二叉树相等当且仅当rootA->data == rootB->data,且A、B的左右子树相等或者左右互换相等

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        if (!root1 && !root2)
            return 1;
        if ((!root1&&root2) || (root1 && !root2))
            return 0;
        if (root1&&root2)
        {
            if (root1->val == root2->val)
            {
                if (flipEquiv(root1->left, root2->left))
                    return flipEquiv(root1->right, root2->right);
                else if (flipEquiv(root1->left, root2->right))
                    return flipEquiv(root1->right, root2->left);
            }
        }
        return 0;
    }
};

 

113th LeetCode Weekly Contest Flip Equivalent Binary Trees

标签:tree node   左右子树   ati   unique   opera   ons   make   output   def   

原文地址:https://www.cnblogs.com/yinghualuowu/p/10055358.html

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