[leetcode] Construct binary tree from inorder and postorder traversal

标签：leetcode

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

postorder 为左 右 中，所以postorder的最后一个为根，取得树根后，在inorder中可以找到左子树与右子树的元素。

问题可分解为，找根，确定左子树与右子树的元素。所以，递归求解。

由于，postorder是从左至右存储根节点的，所以要先构造右子树。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int postindex;
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.size()==0||postorder.size()==0)
            return NULL;
      
        postindex=postorder.size()-1;
        int stat=0;
        int end=postorder.size()-1;
        return foo(inorder,postorder,stat,end);
        
   
     
        
    }
    TreeNode* foo(vector<int> &inorder,vector<int> &postorder,int stat,int end){
        if(stat>end)
            return NULL;
        TreeNode* root=new TreeNode(0);
        root->val=postorder[postindex];
        postindex=postindex-1;
        int indexOfRoot=index(inorder,root->val);
        if(indexOfRoot==-1)
            exit(0);
        if(stat==end)
            return root;
        root->right=foo(inorder,postorder,indexOfRoot+1,end);
        root->left=foo(inorder,postorder,stat,indexOfRoot-1);
        
        return root;
        
    }
    
    int index(vector<int> v,int val){
        int i;
        for(i=0;i<v.size();i++)
            if(v[i]==val)
                return i;
        return -1;
    }
    
};

[leetcode] Construct binary tree from inorder and postorder traversal

标签：leetcode

原文地址：http://lichunyu.blog.51cto.com/8160154/1584340