LeetCode - 106 - 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.

给出一棵二叉树的中根遍历和后根遍历，求这棵树。

基本思路：我们有in和post数组长度都为7，那么我们可以知道post[6]是根。找到post[6]对应in中的位置，假设是in[3]。那么可以得出in[0]~in[2]为左子树，in[4]~in[6]为右子树。同理继续向下递归。

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {return solve(0, postorder.size()-1, 0, inorder.size()-1, inorder, postorder);}TreeNode* solve(int post, int poed, int inst, int ined, vector<int>& inorder, vector<int>& postorder) {if (inst > ined || post > poed) return NULL;TreeNode* cur = new TreeNode(postorder[poed]);int index = 0;for (int i = inst; i <= ined; ++i) {if (inorder[i] != cur->val) continue;index = i;break;}cur->left = solve(post, poed-(ined-index+1), inst, index-1, inorder, postorder);cur->right = solve(post+(index-inst), poed-1, index+1, ined, inorder, postorder);return cur;}
};

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