106. Construct Binary Tree from Inorder and Postorder Traversal

Posted 2020-07-30 积少成多

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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利用:中序+后序遍历

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code:

/**
 * 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:
    ///
    template<typename Iterator>
    TreeNode *help_buildTree_ip(Iterator in_first,Iterator in_last,
        Iterator post_first,Iterator post_last){
        if(in_first == in_last) return nullptr;
        if(post_first == post_last) return nullptr;

        auto val = *prev(post_last);
        TreeNode *root = new TreeNode(val);

        auto in_root_pos = find(in_first,in_last,val);
        auto left_size = distance(in_first,in_root_pos);
        auto post_left_last = next(post_first,left_size);

        root->left = help_buildTree_ip(in_first,in_root_pos,post_first,post_left_last);
        root->right = help_buildTree_ip(next(in_root_pos),in_last,post_left_last,prev(post_last));

        return root;
    }
    TreeNode* buildTree_ip(vector<int> &inorder,vector<int> &postorder){
        return help_buildTree_ip(inorder.begin(),inorder.end(),postorder.begin(),postorder.end());
    }
};