力扣106. 从中序与后序遍历序列构造二叉树

 常规题面，直接维护两种的遍历的下标位置进行递归即可。

/**
 * 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:
    vector<int> inorder, postorder;
    TreeNode* rebuild(int il, int ir, int pl, int pr) {
        if (il > ir || pl > pr) return nullptr;
        int root = postorder[pr];
        int i; // 根结点在当前中序遍历[il,ir]中的位置
        for (i = il; i <= ir; ++i) {
            if (inorder[i] == root) break;
        }
        TreeNode * parent = new TreeNode(root);
        // 根据划分后先序遍历的长度来确定划分后序遍历的长度
        parent -> left = rebuild(il, i - 1, pl, pl + i - 1 - il);
        parent -> right = rebuild(i + 1, ir, pl + i -il, pr - 1);
        return parent;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        this -> inorder = inorder;
        this -> postorder = postorder;
        return rebuild(0, inorder.size() - 1, 0, postorder.size() - 1);
    }
};