Construct Binary Tree from Preorder and Inorder Traversal

文章目录

1. Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

题目

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

Note:

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

思路

根据先序遍历和中序遍历建树，建立根节点，然后划分左子树和右子树。递归

解题

c++ 版

/**
 * 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:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
         return solve(preorder,0,inorder,0,inorder.size()-1);
    }
    TreeNode* solve(vector<int>& preorder,int p,vector<int>& inorder,int low,int high){
        if(low>high)
            return NULL;
        TreeNode* root=new TreeNode(preorder[p]);
        int pos=0;
        for(int i=low;i<=high;++i){
            if(preorder[p]==inorder[i]){
                pos=i;
                break;
            }
        }
        root->left=solve(preorder,p+1,inorder,low,pos-1);
        root->right=solve(preorder,p+pos-low+1,inorder,pos+1,high);
        return root;
    }

};

python版

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {integer[]} preorder
    # @param {integer[]} inorder
    # @return {TreeNode}
    def buildTree(self, preorder, inorder):
        return self.solve(preorder,0,inorder,0,len(inorder)-1)
    
    def solve(self,preorder,p,inorder,low,high):
        if low>high:
            return None
        root=TreeNode(preorder[p])
        pos=0
        for i in range(low,high+1):
            if preorder[p]==inorder[i]:
                pos=i
                break
        root.left=self.solve(preorder,p+1,inorder,low,pos-1)
        root.right=self.solve(preorder,p+pos-low+1,inorder,pos+1,high)
        return root