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94.binary-tree-inorder-traversal.md

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题目地址

https://leetcode.com/problems/binary-tree-inorder-traversal/description/

题目描述

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]
Follow up: Recursive solution is trivial, could you do it iteratively?

思路

递归的方式相对简单,非递归的方式借助栈这种数据结构实现起来会相对轻松。

如果采用非递归,可以用栈(Stack)的思路来处理问题。

中序遍历的顺序为左-根-右,具体算法为:

  • 从根节点开始,先将根节点压入栈

  • 然后再将其所有左子结点压入栈,取出栈顶节点,保存节点值

  • 再将当前指针移到其右子节点上,若存在右子节点,则在下次循环时又可将其所有左子结点压入栈中, 重复上步骤

94.binary-tree-inorder-traversal

(图片来自: https://github.com/MisterBooo/LeetCodeAnimation)

关键点解析

  • 二叉树的基本操作(遍历)

不同的遍历算法差异还是蛮大的

  • 如果非递归的话利用栈来简化操作

  • 如果数据规模不大的话,建议使用递归

  • 递归的问题需要注意两点,一个是终止条件,一个如何缩小规模

  1. 终止条件,自然是当前这个元素是null(链表也是一样)

  2. 由于二叉树本身就是一个递归结构, 每次处理一个子树其实就是缩小了规模, 难点在于如何合并结果,这里的合并结果其实就是left.concat(mid).concat(right), mid是一个具体的节点,left和right递归求出即可

代码

  • 语言支持:JS,C++,Python3, Java

JavaScript Code:

/*
 * @lc app=leetcode id=94 lang=javascript
 *
 * [94] Binary Tree Inorder Traversal
 *
 * https://leetcode.com/problems/binary-tree-inorder-traversal/description/
 *
 * algorithms
 * Medium (55.22%)
 * Total Accepted:    422.4K
 * Total Submissions: 762.1K
 * Testcase Example:  '[1,null,2,3]'
 *
 * Given a binary tree, return the inorder traversal of its nodes' values.
 * 
 * Example:
 * 
 * 
 * Input: [1,null,2,3]
 * ⁠  1
 * ⁠   \
 * ⁠    2
 * ⁠   /
 * ⁠  3
 * 
 * Output: [1,3,2]
 * 
 * Follow up: Recursive solution is trivial, could you do it iteratively?
 * 
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function(root) {
    // 1. Recursive solution
    // if (!root) return [];
    // const left = root.left ? inorderTraversal(root.left) : [];
    // const right = root.right ? inorderTraversal(root.right) : [];
    // return left.concat([root.val]).concat(right);

    // 2. iterative solutuon
    if (!root) return [];
    const stack = [root];
    const ret = [];
    let left = root.left;

    let item = null; // stack 中弹出的当前项

    while(left) {
        stack.push(left);
        left = left.left;
    }
    
    while(item = stack.pop()) {
        ret.push(item.val);
        let t = item.right;

        while(t) {
            stack.push(t);
            t = t.left;     
        }
    }

    return ret;

};

C++ Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<TreeNode*> s;
        vector<int> v;
        while (root != NULL || !s.empty()) {
            for (; root != NULL; root = root->left)
                s.push_back(root);
            v.push_back(s.back()->val);
            root = s.back()->right;
            s.pop_back();
        }
        return v;
    }
};

Python Code:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        """
        1. 递归法可以一行代码完成,无需讨论;
        2. 迭代法一般需要通过一个栈保存节点顺序,我们这里直接使用列表
          - 首先,我要按照中序遍历的顺序存入栈,这边用的逆序,方便从尾部开始处理
          - 在存入栈时加入一个是否需要深化的参数
          - 在回头取值时,这个参数应该是否,即直接取值
          - 简单调整顺序,即可实现前序和后序遍历
        """
        # 递归法
        # if root is None:
        #     return []
        # return self.inorderTraversal(root.left)\
        #     + [root.val]\
        #     + self.inorderTraversal(root.right)
        # 迭代法
        result = []
        stack = [(1, root)]
        while stack:
            go_deeper, node = stack.pop()
            if node is None:
                continue
            if go_deeper:
                # 左右节点还需继续深化,并且入栈是先右后左
                stack.append((1, node.right))
                # 节点自身已遍历,回头可以直接取值
                stack.append((0, node))
                stack.append((1, node.left))
            else:
                result.append(node.val)
        return result

Java Code:

  • recursion
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Integer> res = new LinkedList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        inorder(root);
        return res;
    }
    
    public void inorder (TreeNode root) {
        if (root == null) return;
        
        inorder(root.left);
        
        res.add(root.val);
        
        inorder(root.right);
    }
}
  • iteration
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<> ();
        Stack<TreeNode> stack = new Stack<> ();

        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            res.add(root.val);
            root = root.right;
        }
        return res;
    }
}