LeetCode OJ

题目：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

解题思路：

后序遍历的最后一个元素肯定为根节点；然后在中序遍历的序列中找到该根节点，将中序遍历序列分为左右两部分，对其分别进行递归求解。

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int findIndex(int target, vector<int> &inorder, int begin, int end) {
        for (int i = begin; i <= end; i++) {
            if (inorder[i] == target) {
                return i;
            }
        }
        return -1;
    }
    TreeNode *buildTreeCore(vector<int> &inorder, int in_begin, int in_end, vector<int> &postorder, int post_begin, int post_end) {
        if (post_begin > post_end || in_begin > in_end) return NULL;
        
        TreeNode *root = new TreeNode(postorder[post_end]);
        if (post_begin == post_end) {
            return root;
        } else {
            int index = findIndex(postorder[post_end], inorder, in_begin, in_end);
            int left_nodes = index - in_begin;
            TreeNode *left_root = buildTreeCore(inorder, in_begin, index - 1, postorder, post_begin, post_begin + left_nodes - 1);
            TreeNode *right_root = buildTreeCore(inorder, index + 1, in_end, postorder, post_begin + left_nodes, post_end - 1);
            root->left = left_root;
            root->right = right_root;
            return root;
        }
    }
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        TreeNode *root = buildTreeCore(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
        return root;
    }
};