105. Construct Binary Tree from Preorder and Inorder Traversal

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最新推荐文章于 2021-05-26 18:38:00 发布

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分类专栏： LeetCode

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本文介绍了一种算法，该算法利用给定的先序和中序遍历序列来构建二叉树。通过递归地查找根节点在中序遍历中的位置，可以确定左右子树的范围，进而构建整个二叉树结构。

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105. Construct Binary Tree from Preorder and Inorder Traversal

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

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/**
 * 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) {
        TreeNode* root=build(0,inorder.size()-1,0,preorder.size()-1,preorder,inorder);
        return root;
    }
    TreeNode* build(int inl,int inr,int prel,int prer,vector<int>& preorder, vector<int>& inorder){
    	if(prel>prer) return NULL;
    	TreeNode* root=new TreeNode(preorder[prel]);
    	//root->val=preorder[prel];
    	//root->left=NULL;
    	//root->right=NULL;
    	if(inl==inr) return root;
    	int k;
    	for(k=inl;k<=inr;k++){//根据前序第一个为根找到中序中的根节点
    		if(preorder[prel]==inorder[k]){
    			break;
    		}
    	}
    	root->left=build(inl,k-1,prel+1,prel+(k-inl),preorder,inorder);//递归建立左子树
    	root->right=build(k+1,inr,prel+(k-inl)+1,prer,preorder,inorder);
    	return root;
    }
};