951. Flip Equivalent Binary Trees

Medium

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.

Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.

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.

Example 2:

Input: root1 = [], root2 = []
Output: true

Example 3:

Input: root1 = [], root2 = [1]
Output: false

Constraints:

• The number of nodes in each tree is in the range [0, 100].

• Each tree will have unique node values in the range [0, 99].

解題

Runtime: 0 ms, faster than 100%

Memory Usage: 2.4 MB, less than 77.78%

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flipEquiv(root1 *TreeNode, root2 *TreeNode) bool {
    if root1 == nil && root2 == nil { // 兩個都nil，相等，回傳true
        return true
    }

    if root1 == nil || root2 == nil || root1.Val != root2.Val { 
        // 只有一邊nil或root值不相等，樹必然不相同，回傳false，
        return false
    } 
    
    return (flipEquiv(root1.Left, root2.Left) && flipEquiv(root1.Right, root2.Right)) || (flipEquiv(root1.Left, root2.Right) && flipEquiv(root1.Right, root2.Left))
    // 不用交換樹就相同 || 交換後樹變相同
}