LeetCode（106） Construct Binary Tree from Inorder and Postorder Traversal

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.

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分析

跟上一道题同样的道理。

AC代码

class Solution {
public:
    template <typename Iter>
    TreeNode* make(Iter in_begin, Iter in_end , Iter post_begin, Iter post_end ) {
        if (post_begin == post_end || in_begin == in_end)
            return NULL;
        int size = post_end - post_begin;
        //后序遍历最后一个节点为树的根节点
        TreeNode *root = new TreeNode(*(post_begin + size-1));
        //在中序遍历结果中查找根节点
        Iter iter = find(in_begin, in_end, *(post_begin + size - 1));
        //计算左子树个数
        int count = iter - in_begin;
        if (iter != in_end)
        {
            //则在inOrder中（0 ， count-1）为左子树中序遍历结果（count+1，size-1）为右子树的中序遍历序列
            //在preOrder中（0，count-1）为左子树前序遍历结果（count,size-2）为右子树前序遍历结果
            root->left = make(in_begin, iter , post_begin, post_begin + count);
            //构造右子树
            root->right = make(iter + 1, in_end ,post_begin + count, post_begin + size - 1);
        }
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.empty() || postorder.empty())
            return NULL;
        return make(inorder.begin(), inorder.end() ,postorder.begin(), postorder.end());
    }
};

GitHub测试程序源码