Tree——No.94 Binary Tree Inorder Traversal

Problem：

Given a binary tree, return the inorder traversal of its nodes' values.

Explanation：

中序遍历给定二叉树

My Thinking：

使用常规递归进行中序遍历，不过这里需要添加一个函数，才能将结果列表作为参数传递。

My Solution：

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

Optimum Thinking:

1. 同my thinking
2. 使用栈

Optimum Solution：

(2)

public class Solution {
    public List < Integer > inorderTraversal(TreeNode root) {
        List < Integer > res = new ArrayList < > ();
        Stack < TreeNode > stack = new Stack < > ();
        TreeNode curr = root;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {//从根结点开始将所有左孩子入栈
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();//弹出一个左孩子
            res.add(curr.val);//将其加入列表
            curr = curr.right;//将其右结点继续循环
        }
        return res;
    }
}

标签：Binary,curr,root,Tree,stack,TreeNode,Inorder,List,numlist
来源： https://blog.csdn.net/qq_41773240/article/details/88380828