951. Flip Equivalent Binary Trees(反转等效二叉树)

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.
951. Flip Equivalent Binary Trees(反转等效二叉树)_第1张图片

方法思路

Approach1:recursive

class Solution {
    //Runtime: 2 ms, faster than 100.00%
    //Memory Usage: 36.8 MB, less than 78.85% 
    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)));
    }
}

Approach2:traversal

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);
            }
        }
    }
}

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