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.
Recursive Solution：
/**
* 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) {
// Start typing your Java solution below
// DO NOT write main() function
if(root==null) return true;
return isSymmetric(root.left,root.right);
}
public boolean isSymmetric(TreeNode a, TreeNode b){
if(a==null) return b==null;
if(b==null) return false;
if(a.val!=b.val) return false;
if(!isSymmetric(a.left,b.right)) return false;
if(!isSymmetric(a.right,b.left)) return false;
return true;
}
}
Iterative Solution
/**
* 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) {
// Start typing your Java solution below
// DO NOT write main() function
if(root==null) return true;
LinkedList<TreeNode> l = new LinkedList<TreeNode>(),
r = new LinkedList<TreeNode>();
l.add(root.left);
r.add(root.right);
while(!l.isEmpty() && !r.isEmpty()){
TreeNode temp1=l.poll(),
temp2=r.poll();
if(temp1==null && temp2!=null || temp1!=null && temp2==null)
return false;
if(temp1!=null){
if(temp1.val!=temp2.val) return false;
l.add(temp1.left);
l.add(temp1.right);
r.add(temp2.right);
r.add(temp2.left);
}
}
return true;
}
}
4 comments:
Beautiful recursive solution.
Why if(a == NULL) b= NULL?
because the two lines are like this:
if(a==null) return b==null;
so if a is NULL then we only need to see if b is NULL, we have two scenarios:
1. b is NULL, then it should return TRUE(both a and b are NULL, it is symmetric);
2. b is NOT NULL, then it should return FALSE(a is NULL but b is NOT NULL, it is NOT symmetric);
see the result is exactly same as the line "if(a==null) return b==null;"
this line is a bit hard to understand, maybe we can write it in a way easier to understand, like:
if (a == null && b == null) {
return true;
} else (a == null && b != null) {
return false;
}
but this is a bit verbose too, it is update the coder.
l.add(temp1.left);
what if temp1.left equals to null? I thought add() can not add a null value.
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