今日leetcode（回顾树）

疯狂刷自己以前的题难受

(28条消息) Java实现二叉树的前序，中序，后序遍历(迭代，递归两种方法实现)_征途远方的博客-CSDN博客_java二叉树的先序,中序,后序遍历

 这里用的DFS

/**
 * 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;
 *     }
 * }
 */
class Solution {
  public List> levelOrderBottom(TreeNode root){
        LinkedList> res = new LinkedList<>();
        dfs(root,0,res);
        return res;
    }

    /**
     * DFS实现层序遍历，将结点添加到我们的集合中res
     * @param node
     * @param depth
     * @param res
     */
    public void dfs(TreeNode node,int depth,List> res){
        //1.结束条件
        if(node==null){
            return;
        }
        //2.实现逻辑,当深度==最大res时，res进行扩容
        if(depth==res.size()){
            res.add(0,new ArrayList<>());
        }

        //3.添加每层值
        res.get(res.size()-depth-1).add(node.val);
        
        //4.递归,深度+1，往上走
        dfs(node.left,depth+1,res);
        dfs(node.right,depth+1,res);
    }


}

/*
// Definition for a Node.
class Node {
    public int val;
    public List children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List preorder(Node root) {
        List res=new ArrayList<>();
        digui(root,res);
        return res;
    }
    public void digui(Node root,List res){
        //1.结束条件,递归到极限就没有头节点
        if(root==null){
          return;
        }
        //2.添加头节点
        res.add(root.val);
        //3.递归
        for(Node children:root.children){
           digui(children,res);
        }
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public int maxDepth(TreeNode root) {
       //1.递归结束条件
       if(root==null){
           return 0;
       }
        int Left = maxDepth(root.left);
        int Right = maxDepth(root.right);
       //2.递归
       return Left>Right?Left+1:Right+1;
    }
}