标签：tree order instead enter empty offer initial point efi
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
M1: recursive
time: O(n) -- n: total # of nodes in the tree, space: O(height)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean isSymmetric(TreeNode root) { if(root == null || root.left == null && root.right == null) { return true; } return isSymmetric(root.left, root.right); } public boolean isSymmetric(TreeNode t1, TreeNode t2) { if(t1 == null && t2 == null) { return true; } if(t1 == null || t2 == null) { return false; } if(t1.val != t2.val) { return false; } return isSymmetric(t1.left, t2.right) && isSymmetric(t1.right, t2.left); } }
M2: iterative
Instead of recursion, we can also use iteration with the aid of a queue. Each two consecutive nodes in the queue should be equal, and their subtrees a mirror of each other. Initially, the queue contains root and root. Then the algorithm works similarly to BFS, with some key differences. Each time, two nodes are extracted and their values compared. Then, the right and left children of the two nodes are inserted in the queue in opposite order. The algorithm is done when either the queue is empty, or we detect that the tree is not symmetric (i.e. we pull out two consecutive nodes from the queue that are unequal).
time: O(n), space: O(n) -- worst case need to store n nodes in the queue
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean isSymmetric(TreeNode root) { Queue<TreeNode> q = new LinkedList<>(); q.offer(root); q.offer(root); while(!q.isEmpty()) { TreeNode t1 = q.poll(); TreeNode t2 = q.poll(); if(t1 == null && t2 == null) { continue; } if(t1 == null || t2 == null) { return false; } if(t1.val != t2.val) { return false; } q.add(t1.left); q.add(t2.right); q.add(t1.right); q.add(t2.left); } return true; } }
标签：tree order instead enter empty offer initial point efi
原文地址：https://www.cnblogs.com/fatttcat/p/10188694.html
