LeetCode: Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
地址：https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
算法：根据中序和后序构造出二叉树。细心一点应该不会错。代码：
1 /**
2 * Definition for binary tree
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10 class Solution {
11 public:
12 TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
13 if(inorder.empty()) return NULL;
14 return subBuildTree(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);
15 }
16 TreeNode *subBuildTree(vector<int> &inorder, vector<int> &postorder, int inbegin, int inend, int postbegin, int postend){
17 TreeNode *root = new TreeNode(postorder[postend]);
18 if(inbegin == inend && postbegin == postend){
19 return root;
20 }
21 int key = postorder[postend];
22 int i = inbegin;
23 while(i <= inend && inorder[i] != key) ++i;
24 if(i > inbegin){
25 root->left = subBuildTree(inorder,postorder,inbegin,i-1,postbegin,postbegin+i-1-inbegin);
26 }
27 if(inend > i){
28 root->right = subBuildTree(inorder,postorder,i+1,inend,postbegin+i-inbegin,postend-1);
29 }
30 return root;
31 }
32 };
