Invert Binary Tree

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.

分析：翻转二叉树

思路1：

使用递归的方式，时间复杂度为o(n)，空间复杂度为o(n)

JAVA CODE

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null||(root.left==null&&root.right==null))
            return root;
        if(root.left!=null)
            invertTree(root.left);
        if(root.right!=null)
            invertTree(root.right);
        TreeNode tempL = root.left;
        TreeNode tempR = root.right;
        root.left = tempR;
        root.right = tempL;
        return root;
    }
}

 

思路2：

使用迭代的方式，时间复杂度为o(n)，空间复杂度为o(1)

JAVA CODE

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