226. Invert Binary Tree

题目：

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

链接： http://leetcode.com/problems/invert-binary-tree/ 

题解：

其实我还加这哥们LinkedIn了...后来他被Apple录取了，挺好的。 使用Recursive比较快就能写出来。

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 TreeNode invertTree(TreeNode root) {
        if(root == null)
            return root;
        TreeNode left = root.left;
        root.left = invertTree(root.right);
        root.right = invertTree(left);
        
        return root;
    }
}

 

二刷：

使用递归求解比较方便。先保存左子树，，再更新左子树和右子树，最后返回root。

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 TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        TreeNode tmpLeft = root.left;
        root.left = invertTree(root.right);
        root.right = invertTree(tmpLeft);
        return root;
    }
}

 

使用一个stack来迭代

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        LinkedList<TreeNode> stack = new LinkedList<>();
        stack.addLast(root);
        while (stack.size() > 0) {
            TreeNode node = stack.pollLast();
            if (node != null) {
                TreeNode tmp = node.left;
                node.left = node.right;
                node.right = tmp;
                stack.addLast(node.left);
                stack.addLast(node.right);    
            }
        }
        return root;
    }
}

 

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 TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        TreeNode tmpLeft = root.left;
        root.left = invertTree(root.right);
        root.right = invertTree(tmpLeft);
        return root;
    }
}

 

使用一个stack来模拟系统栈， 用Queue也可以。 以后我决定在面试里就写stack，不用Doubly LinkedList了。这样表达更清晰。自己清楚Stack是继承自vector，并且是depricated就可以了，在production上不用, 面试时提一下。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (node != null) {
                TreeNode tmpLeft = node.left;
                node.left = node.right;
                node.right = tmpLeft;
                stack.push(node.left);
                stack.push(node.right);
            }
        }
        return root;
    }
}

 

用Queue的

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return root;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            if (node != null) {
                q.offer(node.left);
                q.offer(node.right);
                TreeNode tmp = node.left;
                node.left = node.right;
                node.right = tmp;
            }
        }
        return root;
    }
}

 

 

Reference:

https://leetcode.com/discuss/40001/straightforward-dfs-recursive-iterative-bfs-solutions

https://leetcode.com/discuss/40051/3-4-lines-python