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[Leetcode]106. Construct Binary Tree from Inorder and Postorder Traversal(C++)

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题目描述

题目链接：106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

例子

见题目描述

Note

You may assume that duplicates do not exist in the tree.

解题思路

思路和 105 类似，区别在于要从 postorder 中的尾部获取根节点再从 inorder 获取左右子树长度，代码如下：

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
#include <unordered_map>
#include <vector>

class Solution {
public:
    TreeNode* buildTree(std::vector<int>& inorder,
                        std::vector<int>& postorder) {
        std::unordered_map<int, int> inorder_map;
        for (size_t i = 0; i < inorder.size(); ++i) {
            inorder_map[inorder[i]] = i;
        }
        return buildTree(postorder, inorder, 0, postorder.size() - 1, 0,
                         inorder.size() - 1, inorder_map);
    }

private:
    TreeNode* buildTree(const std::vector<int>& postorder,
                        const std::vector<int>& inorder, int postorder_begin,
                        int postorder_end, int inorder_begin, int inorder_end,
                        std::unordered_map<int, int>& inorder_map) {
        if (postorder_end < postorder_begin || inorder_end < inorder_begin)
            return nullptr;
        TreeNode* root = new TreeNode(postorder[postorder_end]);
        int root_idx = inorder_map[root->val];
        int left_len = root_idx - inorder_begin;
        int right_len = inorder_end - root_idx;
        root->left = buildTree(postorder, inorder, postorder_begin,
                               postorder_begin + left_len - 1, inorder_begin,
                               root_idx - 1, inorder_map);
        root->right = buildTree(postorder, inorder, postorder_begin + left_len,
                                postorder_end - 1, root_idx + 1, inorder_end,
                                inorder_map);

        return root;
    }
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(n)

GitHub 代码同步地址： 106.ConstructBinaryTreeFromInorderAndPostorderTraversal.cpp

其他题目：
GitHub: Leetcode-C++-Solution
博客： Leetcode-Solutions

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