翻转二叉树（简单）

2020年6月1日

题目来源：力扣

解题

递归

深度遍历节点，交换两个子节点的内容

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

迭代
二叉树的迭代经常使用队列，节点入队列，把节点的两个子节点交换后，再把两个子节点入队列

/**
 * 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;
        Queue<TreeNode> q=new LinkedList<>();
        q.add(root);
        while(!q.isEmpty()){
            TreeNode cur=q.poll();
            TreeNode tmp=cur.left;
            cur.left=cur.right;
            cur.right=tmp;
            if(cur.left!=null) q.add(cur.left);
            if(cur.right!=null) q.add(cur.right);
        }
        return root;
    }
}