[LeetCode]106. Construct Binary Tree from Inorder and Postorder Traversal

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本文介绍了一种根据二叉树的中序和后序遍历结果重构该二叉树的方法。通过递归地确定根节点及其左右子树，最终完成整个二叉树的构建。

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.

/**
 * 和之前使用中序和前序遍历重建的题一样，可以直接使用索引，避免反复创建vector，这里就省略不实现了...
 */
/**
 * 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() || postorder.empty())
            return NULL;
        TreeNode* root = new TreeNode(*(postorder.end()-1));
        int i = 0;
        for(; i < inorder.size(); ++i){
            if(inorder[i] == root->val)
                break;
        }
        if(i > 0){
            vector<int> in_left = vector<int>(inorder.begin(), inorder.begin()+i);
            vector<int> post_left = vector<int>(postorder.begin(), postorder.begin()+i);
            root->left = buildTree(in_left, post_left);
        }
        if(i < inorder.size()-1){
            vector<int> in_right = vector<int>(inorder.begin()+i+1, inorder.end());
            vector<int> post_right = vector<int>(postorder.begin()+i, postorder.end()-1);
            root->right = buildTree(in_right, post_right);
        }
        return root;
    }
};