Alibi

My Life is Getting Better

Leetcode[101] Symmetric Tree

07 Nov 2015

###Task1 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

###Python ####Recursive

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        return self.isValid(root.left, root.right)
    def isValid(self, left, right):
        if not left and not right:
            return True
        if not left or not right:
            return False
        if left.val != right.val:
            return False
        return self.isValid(left.left, right.right) and self.isValid(left.right, right.left)

####Iterative

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        leftQ = collections.deque()
        rightQ = collections.deque()
        leftQ.append(root.left)
        rightQ.append(root.right)
        while len(leftQ) > 0 or len(rightQ) > 0:
            l = leftQ.popleft()
            r = rightQ.popleft()
            if not l and not r:
                continue
            if not l or not r:
                return False
            if l.val != r.val:
                return False
            leftQ.append(l.left)
            leftQ.append(l.right)
            rightQ.append(r.right)
            rightQ.append(r.left)
        return True

###Java

/**
 * Definition for a binary tree node.
 * 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;
        }
        if (root.left == null && root.right == null) {
            return true;
        }
        if (root.left == null || root.right == null) {
            return false;
        }

        Queue<TreeNode> leftQueue = new LinkedList<TreeNode>();
        Queue<TreeNode> rightQueue = new LinkedList<TreeNode>();
        leftQueue.offer(root.left);
        rightQueue.offer(root.right);

        while (!leftQueue.isEmpty() && !rightQueue.isEmpty()) {
            TreeNode left = leftQueue.poll();
            TreeNode right = rightQueue.poll();
            if (left == null && right == null) {
                continue;
            }
            if (left == null || right == null) {
                return false;
            }
            if (left.val != right.val) {
                return false;
            } else {
                leftQueue.offer(left.left);
                leftQueue.offer(left.right);
                rightQueue.offer(right.right);
                rightQueue.offer(right.left);
            }
        }

        return true;
    }
}

###Points

Related Posts