Leetcode 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.

confused what

"{1,#,2,3}"

means?


OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

.

解题思路：

1. 递归

2. iterative,建立两个queue, 一个从左往右，一个从右往左，镜像比较

仔细考虑每种情况，很容易漏掉。

Java code :

recursion:

public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root == null) {
return true;
}
return isSymmetric(root.left, root.right);
}

public boolean isSymmetric(TreeNode l, TreeNode r) {
if(l == null && r == null){
return true;
}else if(l == null || r == null) {
return false;
}
if(l.val != r.val ){
return false;
}
if(!isSymmetric(l.left,r.right)) {
return false;
}
if(!isSymmetric(l.right,r.left)) {
return false;
}
return true;
}
}

iterative

public class Solution {
public boolean isSymmetric(TreeNode root) {
//use iterative
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.isEmpty() && !right.isEmpty()) {
TreeNode l = left.remove();
TreeNode r = right.remove();
if(l== null && r== null) {
continue;
}else if(l == null || r == null){
return false;
}

if(l.val != r.val) {
return false;
}else {
left.add(l.left);
left.add(l.right);
right.add(r.right);
right.add(r.left);
}
}
return true;
}
}

Reference:

1. http://www.programcreek.com/2014/03/leetcode-symmetric-tree-java/