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  • leetcode-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.

    /**
     * Definition for binary tree
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
        TreeNode *build(vector<int>&in,int instart,int inend,vector<int>&post,int poststart,int postend)
        {
            if(postend<poststart||inend<instart||(postend-poststart)!=(inend-instart))
                return NULL;
            TreeNode *head = new TreeNode(post[postend]);
            int rootpos;
            for(int i=0;i<=inend-instart;i++)
                if(in[i+instart]==post[postend])
                {
                    rootpos = i;
                    break;
                }
                
            head->left  = build(in,instart,instart+rootpos-1,post,poststart,poststart+rootpos-1);
            head->right = build(in,instart+rootpos+1,inend,post,poststart+rootpos,postend-1);
            return head;
        }
        TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) 
        {
            int postsize = postorder.size();
            int insize   = inorder.size();
            TreeNode *head = build(inorder,0,insize-1,postorder,0,postsize-1);
            return head;
        }
    };


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  • 原文地址:https://www.cnblogs.com/vintion/p/4116858.html
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