106. Construct Binary Tree from Inorder and Postorder Traversal

目录

• 题目描述
• 代码实现

题目描述

Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

代码实现

class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {

    	int m = inorder.size();
    	int n = postorder.size();
    	if(m!=n || m==0)
    		return nullptr;
    	return buildTreeCore(inorder,0,n-1,postorder,0,n-1);
        
    }

    TreeNode *buildTreeCore(vector<int> &inorder,int in_l,int in_r,vector<int> &postorder,int post_l,int post_r)
    {      
        if(in_l>in_r)
            return nullptr;                           
    	TreeNode * root = new TreeNode(postorder[post_r]);
    	root->left = nullptr;
    	root->right = nullptr;

    	int theRoot = in_l;
    	while(inorder[theRoot]!=root->val)theRoot++;

        root->left = buildTreeCore(inorder,in_l,theRoot-1,postorder,post_l,post_l+theRoot-in_l-1);
        root->right = buildTreeCore(inorder,theRoot+1,in_r,postorder,post_l+theRoot-in_l,post_r-1);

        return root;

    }
};