106-construct-binary-tree-from-inorder-and-postorder-traversal.md

106. Construct Binary Tree from Inorder and Postorder Traversal - 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。

注意:
你可以假设树中没有重复的元素。

例如，给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

---

题目标签：Tree / Depth-first Search / Array

题目链接：LeetCode / LeetCode中国

题解

Language	Runtime	Memory
cpp	16 ms	1.3 MB

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
static auto _ = [](){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return 0;
}();

class Solution {
public:
    void build(vector<int>& in, vector<int>& post, int il, int ir, int pl, int pr, TreeNode*& ret) {
        if (il >= ir || pl >= pr) {
            return;
        }
        ret = new TreeNode(post[pr-1]);
        int x = 0;
        for (int i=il; i<ir; ++i) {
            if (in[i] == post[pr-1]) {
                x = i - il;
                break;
            }
        }
        build(in, post, il+x+1, ir, pl+x, pr-1, ret->right);
        build(in, post, il, il+x, pl, pl+x, ret->left);
    }
    
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        TreeNode* root = nullptr;
        build(inorder, postorder, 0, inorder.size(), 0, postorder.size(), root);
        return root;
    }
};