101. Symmetric Tree

题目：

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? > read more on how binary tree is serialized on OJ.

 

Hide Tags

 Tree Depth-first Search

 

链接：http://leetcode.com/submissions/detail/11328058/

题解：

DFS。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null)
            return true;
        else
            return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right){
        if(left == null && right == null)
            return true;
        if(left != null && right != null && left.val == right.val)
            return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
        else
            return false;
    }
}

 

Update:

/**
 * 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;
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if(left == null && right == null)
            return true;
        else if (left == null || right == null)
            return false;
        else if(left.val != right.val)
            return false;
        else
            return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
    }
}

 

二刷：

还是使用recursive方法比较简单。就是做一个辅助方法，比较root.left和root.right，以及接下来的root.left的子节点和root.right的子节点。

其他方法也可以用in-order traversal， level-order traversal等等。

Java:

Time Complexity - O(n)， Space Complexity - O(n)。

/**
 * 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;
        }
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == null || right == null) {
            return left == null && right == null;
        }
        if (left.val != right.val) {
            return false;
        }
        return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
    }
}

 

三刷:

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;
        }
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == null || right == null) {
            return left == right;
        }
        if (left.val == right.val) {
            return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
        }
        return false;
    }
}

 

 

Reference:

https://leetcode.com/discuss/19859/slim-java-solution