标签：init win ati off elf poi for roo while
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.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Pair{ TreeNode x; TreeNode y; public Pair(TreeNode x, TreeNode y){ this.x = x; this.y = y; } } public class Solution { // recursive // public boolean isSymmetric(TreeNode root) { // if(root == null) // return true; // return Symmetric(root.left, root.right); // } // public boolean Symmetric(TreeNode left, TreeNode right){ // if(left == null || right == null){ // return left == right; // } // if(left.val != right.val){ // return false; // } // return Symmetric(left.left, right.right) && Symmetric(left.right, right.left); // } public boolean isSymmetric(TreeNode root) { if(root == null || root.left == null && root.right == null) return true; Queue<Pair> queue = new LinkedList<Pair>(); queue.offer(new Pair(root.left, root.right)); while(!queue.isEmpty()){ Pair values= queue.poll(); if(values.x == null && values.y == null) continue; if(values.x == null || values.y == null || values.x.val != values.y.val) return false; if(values.x != null && values.y != null){ queue.offer(new Pair(values.x.left, values.y.right)); queue.offer(new Pair(values.x.right, values.y.left)); } } return true; } }
标签：init win ati off elf poi for roo while
原文地址：http://www.cnblogs.com/joannacode/p/5998968.html
