144. 二叉树的前序遍历

94. 二叉树的中序遍历

145. 二叉树的后序遍历

题目描述

给定一个二叉树的根节点 root ，返回它的三序遍历

输入输出

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2
3
4
5
6
7
8
9
10
    1
     \
     2
    /  
   3

输入：root = [1,null,2,3]
前序 输出：[1,2,3]
中序 输出：[1,3,2]
后序 输出：[1,3,2]

java实现

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/**
 * 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;
 *     }
 * }
 */
//前序
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        preorder(root, result);
        return result;
    }
    void preorder(TreeNode root, List<Integer> res){
        if(root == null) return;
        res.add(root.val);
        preorder(root.left, res);
        preorder(root.right, res);
    }
}

// 中序
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        inorder(root, res);
        return res;
    }
    void inorder(TreeNode root, List<Integer> res){
        if(root == null){
            return;
        }
        inorder(root.left, res);
        res.add(root.val);
        inorder(root.right, res);
    }
}

//后序
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        postorder(root, res);
        return res;
    }
    void postorder(TreeNode root, List<Integer> res){
        if(root == null) return;
        postorder(root.left, res);
        postorder(root.right, res);
        res.add(root.val);
    }
}