leecode106. 从中序与后序遍历序列构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树。

C++:

这里的root->right和root->left的顺序没有影响，先写root->left或者root->right都可以。

/**
 * 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 {
private:
    unordered_map<int, int> map;
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int n = inorder.size();
        for (int i = 0; i<n; i++){
            map[inorder[i]] = i;
        }
        return mybuildTree(inorder, postorder, 0, n-1, 0, n-1);
    }
    TreeNode* mybuildTree(const vector<int>& inorder, const vector<int>& postorder, int in_left, int in_right, int pos_left, int pos_right){
        if (in_left>in_right || pos_left>pos_right) return nullptr;
        TreeNode* root = new TreeNode(postorder[pos_right]);
        int in_root = map[postorder[pos_right]];
        int left_subtree_size =  in_root - in_left -1;
        int right_subtree_size = in_right - in_root -1;
        
        root->left = mybuildTree(inorder, postorder, in_left, in_root-1, pos_left, pos_left + left_subtree_size);
        root->right = mybuildTree(inorder, postorder, in_root+1, in_right, pos_right-right_subtree_size-1, pos_right-1);
        
        return root;
    }
};

C++官方：

自己按照官方的思路写了一下，一直报错找不到原因，后来发现在buildTree中的pos_idx前面加了int，是不是因为前面定义了一次int pos_idx，后面又定义int pos_idx = (int) inorder.size()-1,相当于重新分配内存了？，由于没学过C++，具体原因不清楚。

这里root->left和root->right的顺序不能变，必须是先构建右子树，再构建左子树，即先root->righ=...再root->left =...。

class Solution {
    int pos_idx;
    unordered_map<int, int> map;
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        pos_idx = inorder.size()-1;
        for(int i =0; i<inorder.size(); i++){
            map[inorder[i]] = i;
        }
        return mybuildTree(inorder, postorder, 0, inorder.size()-1);
    }
    TreeNode* mybuildTree(const vector<int>& inorder, const vector<int>& postorder, int in_left, int in_right){
        if (in_left>in_right) return nullptr;

        TreeNode* root = new TreeNode(postorder[pos_idx]);
        int in_root = map[postorder[pos_idx]];
        pos_idx--;
        
        root->right = mybuildTree(inorder, postorder, in_root+1, in_right);
        root->left = mybuildTree(inorder, postorder, in_left, in_root-1);

        return root;
    }
};

Python：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
        if len(inorder) == 0:
            return None;
        root_val = postorder[-1]
        root = TreeNode(root_val)
        idx_root = inorder.index(root_val)
        root.left = self.buildTree(inorder[:idx_root], postorder[:idx_root])
        root.right = self.buildTree(inorder[idx_root+1:], postorder[idx_root:-1])
        return root