951.Flip-Equivalent-Binary-Trees

951. Flip Equivalent Binary Trees

题目地址

https://leetcode.com/problems/flip-equivalent-binary-trees/

题目描述

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:
Each tree will have at most 100 nodes.
Each value in each tree will be a unique integer in the range [0, 99].

代码

Approach #1 Recursion

Time: O(mini(N1, N2)) && Space: O(min(H1, H2))

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

Approach 2: Canonical Traversal

Time: O(N1 + N2)

class Solution {
  public boolean flipEquiv(TreeNode root1, TreeNode root2) {
    List<Integer> vals1 = new ArrayList();
    List<Integer> vals2 = new ArrayList();
    dfs(root1, vals1);
    dfs(root2, vals2);
    return vals1.equals(vals2);
  }

  public void dfs(TreeNode node, List<Integer> vals) {
    if (node != null) {
      vals.add(node.val);
      int L = node.left != null ? node.left.val : -1;
      int R = node.right != null ? node.right.val : -1;

      if (L < R) {
        dfs(node.left, vals);
        dfs(node.right, vals);
      }  else {
        dfs(node.right, vals);
        dfs(node.left. vals);
      }

      vals.add(null);
    }
  }
}