LeetCode | 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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

 

//递归解法
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        
        if(root == null){
            return null;
        }
        
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        
        invertTree(root.left);
        invertTree(root.right);
        
        return root;
    }
}

 

//迭代解法
//思路就是利用一个容器作为缓冲，每次从容器中取出一个node，交换它的left与right，再将子节点放入容器
//关键是要保证每个node的left与right都能被交换，而至于容器用queue还是stack都是无所谓的，只要保证容器
//内的node顺序不乱就行
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        
        if(root == null){
            return null;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        
        while(!queue.isEmpty()){
            TreeNode curNode = queue.remove();   //返回并移除队头
            TreeNode temp = curNode.left;
            curNode.left = curNode.right;
            curNode.right = temp;                //交换队头的left与right
            
            if(curNode.left != null){            //再把left与right均入队
                queue.add(curNode.left);
            }
            if(curNode.right != null){           //此处两个入队的顺序是无关紧要的
                queue.add(curNode.right);        //主要就是每次从队中取出一个node，交换它的left与right
            }                                    //然后在把他的子节点也入队，实现类似递归的过程
        }
        
        return root;
    }
}

 

public class Solution {
    public TreeNode invertTree(TreeNode root) {
        
        if(root == null){
            return null;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        
        while(!stack.isEmpty()){
            TreeNode curNode = stack.pop();
            TreeNode temp = curNode.left;
            curNode.left = curNode.right;
            curNode.right = temp;
            
            if(curNode.left != null){
                stack.push(curNode.left);
            }
            if(curNode.right != null){
                stack.push(curNode.right);
            }
        }
        
        return root;
    }
}