二叉树的中序遍历

题意

返回二叉树中序遍历的节点值序列。

思路

递归：左-根-右。迭代：显式栈模拟递归，先把左链全部入栈，弹出访问后转向右子树。

代码

cpp

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65/**
 * Author by Rainboy
 */
#include <bits/stdc++.h>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left, *right;

    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> ans;
        stack<TreeNode *> st;
        auto cur = root;
        while (cur || !st.empty()) {
            while (cur) {
                st.push(cur);
                cur = cur->left;
            }
            cur = st.top();
            st.pop();
            ans.push_back(cur->val);
            cur = cur->right;
        }
        return ans;
    }
};

TreeNode *build(istream &in, int n) {
    if (!n)
        return nullptr;
    vector<TreeNode *> nodes(n);
    for (int i = 0, v; i < n; i++) {
        in >> v;
        if (v != -1)
            nodes[i] = new TreeNode(v);
    }
    for (int i = 0; i < n; i++) {
        if (!nodes[i])
            continue;
        int l = 2 * i + 1, r = 2 * i + 2;
        if (l < n)
            nodes[i]->left = nodes[l];
        if (r < n)
            nodes[i]->right = nodes[r];
    }
    return nodes[0];
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    auto root = build(cin, n);
    auto v = Solution().inorderTraversal(root);
    for (int x : v)
        cout << x << ' ';
    return 0;
}

python

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49#!/usr/bin/env python3
from typing import List, Optional


class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ans, stack = [], []
        cur = root
        while cur or stack:
            while cur:
                stack.append(cur)
                cur = cur.left
            cur = stack.pop()
            ans.append(cur.val)
            cur = cur.right
        return ans


def build(arr):
    if not arr:
        return None
    nodes = [TreeNode(v) if v != -1 else None for v in arr]
    for i in range(len(nodes)):
        if nodes[i] is None:
            continue
        l, r = 2 * i + 1, 2 * i + 2
        if l < len(nodes):
            nodes[i].left = nodes[l]
        if r < len(nodes):
            nodes[i].right = nodes[r]
    return nodes[0]


def main() -> None:
    n = int(input())
    a = list(map(int, input().split()))
    root = build(a)
    print(*Solution().inorderTraversal(root))


if __name__ == "__main__":
    main()

复杂度

时间复杂度：O(n)。
空间复杂度：O(n) 递归栈/显式栈。

总结

中序迭代栈是"沿左链入栈 → 弹出访问 → 转向右子树"三部曲，是后续很多树操作的基础。

二叉树的中序遍历

目录

题意

思路

代码

复杂度

总结