LeetCode 51 Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

     1   /   \  2     2 / \   / \3   4 4   3

But the following is not:

   1  / \ 2   2  \   \   3   3

NOTE：

Bonus points if you could solve it both recursively and iteratively.

分析：

递归解法就是分别判断左右子树是否对称，即左子树的左子树和右子树的右子树应该是对称的，左子树的右子树和右子树的左子树应该是对称的。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isSymmetric(TreeNode root) {        if(root==null) return true;        return isSymmetric(root.left, root.right);    }    private boolean isSymmetric(TreeNode left, TreeNode right){        if(left==null) return right==null;        if(right==null) return false;        if(left.val != right.val) return false;        if(!isSymmetric(left.left, right.right)) return false;        if(!isSymmetric(left.right, right.left)) return false;                return true;    }}

迭代解法

就是分别对左右子树进行层序遍历，对于每一层，左子树从左向右遍历，右子树从右向左遍历，保证左右队列里每一个元素都相等，则是对称的。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isSymmetric(TreeNode root) {                if(root == null) return true;        Queue<TreeNode> left = new LinkedList<TreeNode>();        Queue<TreeNode> right = new LinkedList<TreeNode>();        left.add(root.left);        right.add(root.right);        while(left.size()>0 && right.size()>0){            TreeNode l = left.remove();            TreeNode r = right.remove();            if((l == null && r != null) || (l != null && r == null))                return false;            if(l != null && r != null){                if(l.val != r.val) return false;                left.add(l.left);                left.add(l.right);                right.add(r.right);                right.add(r.left);            }        }        return true;    }}