给你一棵二叉树的根节点 root
,翻转这棵二叉树,并返回其根节点。
示例 1:
输入:root = [4,2,7,1,3,6,9] 输出:[4,7,2,9,6,3,1]
示例 2:
输入:root = [2,1,3] 输出:[2,3,1]
示例 3:
输入:root = [] 输出:[]/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ // 使用栈实现二叉树的翻转 class Solution { public TreeNode invertTree(TreeNode root) { if (root == null) { return null; } Stack<TreeNode> stack = new Stack<>(); stack.push(root); while (!stack.isEmpty()) { TreeNode current = stack.pop(); TreeNode temp = current.left; current.left = current.right; current.right = temp; if (current.left != null) { stack.push(current.left); } if (current.right != null) { stack.push(current.right); } } return root; } } 递归 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;
栈
} } 广度遍历 迭代 class Solution {
public TreeNode invertTree(TreeNode root) { if (root == null) { return null; } // 将二叉树中的节点逐层放入队列中,再迭代处理队列中的元素 Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.add(root); while (!queue.isEmpty()) { // 每次都从队列中拿一个节点,并交换这个节点的左右子树 // 先进先出 TreeNode tmp = queue.poll(); TreeNode left = tmp.left; tmp.left = tmp.right; tmp.right = left; // 如果当前节点的左子树不为空,则放入队列等待后续处理 if (tmp.left != null) { queue.add(tmp.left); } // 如果当前节点的右子树不为空,则放入队列等待后续处理 if (tmp.right != null) { queue.add(tmp.right); } } // 返回处理完的根节点 return root; } }
标签:current,right,TreeNode,力扣,二叉树,226,null,root,left From: https://www.cnblogs.com/JavaYuYin/p/18014396