106. 从中序与后序遍历序列构造二叉树

106. 从中序与后序遍历序列构造二叉树

给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

 

示例 1:

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]

示例 2:

输入：inorder = [-1], postorder = [-1]
输出：[-1]

 

提示:

• 1 <= inorder.length <= 3000
• postorder.length == inorder.length
• -3000 <= inorder[i], postorder[i] <= 3000
• inorder 和 postorder 都由 不同 的值组成
• postorder 中每一个值都在 inorder 中
• inorder 保证是树的中序遍历
• postorder 保证是树的后序遍历

解题思路：

根据中序遍历节点值和位置索引的map可以快速找到子树根节点的位置，结合中序和后序数组的边界可以找到左右子树的范围

 

 

 

/**
 * 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) {}
 * };
 */
class Solution {
public:
    unordered_map<int, int> map; // key->根节点值,value->根节点位置索引
    TreeNode *dfs(vector<int> &inorder, vector<int> &postorder, int iLeft, int iRight, int pLeft, int pRight) {
        if (iLeft > iRight || pLeft > pRight) {
            return nullptr;
        }
        int k = map[postorder[pRight]] - iLeft;
        TreeNode *subRoot = new TreeNode(postorder[pRight]);
        subRoot->left = dfs(inorder, postorder, iLeft, iLeft + k - 1, pLeft, pLeft + k - 1);
        subRoot->right = dfs(inorder, postorder, iLeft + k + 1, iRight, pLeft + k, pRight - 1);
        return subRoot;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.empty() || postorder.empty()) {
            return nullptr;
        }
        if (inorder.size() != postorder.size()) {
            return nullptr;
        }
        for (unsigned int i = 0; i < inorder.size(); i++) {
            map[inorder[i]] = i;
        }
        return dfs(inorder, postorder, 0, inorder.size() -1, 0, postorder.size() - 1);
    }
};