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.

题意：根据中序序列和后序序列，构建出一颗二叉树。

思路：先找根，再找左树的范围，再找右树的范围，分别构建左树右树。

/** * 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) {if (inorder.empty())return NULL;return build(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);}private:TreeNode* build(vector<int>& inorder, int i1, int i2, vector<int>& postorder, int p1, int p2){if (i1 > i2 || p1 > p2)return NULL;TreeNode* root = new TreeNode(postorder[p2]);vector<int>::iterator it = find(inorder.begin() + i1, inorder.begin() + i2, postorder[p2]);int imid = it - inorder.begin();int leftSize = imid - i1;root->left = build(inorder, i1, imid-1, postorder, p1, p1 + leftSize-1);root->right = build(inorder, imid + 1, i2, postorder, p1 + leftSize, p2 - 1);return root;}};