226. 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.

方法1:recursive

/**
 * 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 null;
        TreeNode left = root.left;
        TreeNode right = root.right;
        root.left = invertTree(right);
        root.right = invertTree(left);
        return root;
    }
}

但是树太深会有stack overflow的风险

方法2: BFS - Stack

/**
 * 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 null;
        Deque stack = new LinkedList<>();
        
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            TreeNode left = node.left;
            node.left = node.right;
            node.right = left;
            if(node.left!=null){
                stack.push(node.left);
            }
            if(node.right!=null){
                stack.push(node.right);
            }
        }
        
    }
}