Monday, April 29, 2019

LeetCode[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.
Flipped Trees Diagram

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].
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/**
 * 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 true;
        if(!root1 || !root2) return false;
        if(root1->val != root2->val) return false;
        
        if(flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right)) return true;
        if(flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left)) return true;
        
        return false;
    }
};


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//Java
/**
 * 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 true;
        if(root1==null || root2==null) return false;
        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));
    }
}
at

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