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
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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
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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
(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
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如果非递归的话利用栈来简化操作
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如果数据规模不大的话，建议使用递归
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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
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终止条件，自然是当前这个元素是null（链表也是一样）
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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; } }LeetCode题解145.binary-tree-postorder-traversal
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