Construct Binary Tree from Preorder and Inorder Traversal
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一、问题描述
Given preorder and inorder traversal of a tree, construct the binary tree.
二、思路
二叉树的题目用递归做是最便捷的,本题也不例外。
递归的退出条件是:
if(pre_begin > preorder.size() - 1 || in_begin > in_end) return NULL;
最关键的是在先序遍历中找出当前节点的值,即root -> val。还有我们在循环中加入break,可以使我们的运行时间从20ms减少到16ms。
在执行递归的时:
左子树在先序遍历指向下一个节点,即左子树的起始位置,在中序遍历中的范围是(in_begin,index - 1);
右子树在先序遍历指向是当前位置减去中序遍历开始的位置,即右子树的起始位置,中序遍历中的范围是(index + 1, in_end);
三、代码
/** * 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 creatTree(0, 0, inorder.size() - 1, preorder, inorder); } TreeNode* creatTree(int pre_begin, int in_begin, int in_end, vector<int>& preorder, vector<int>& inorder){ if(pre_begin > preorder.size() - 1 || in_begin > in_end) return NULL; TreeNode *root = new TreeNode(preorder[pre_begin]); int index = 0; for (int i = in_end; i >= in_begin; --i) { if (inorder[i] == root -> val) { index = i; break; } } root -> left = creatTree(pre_begin + 1, in_begin, index - 1, preorder, inorder); root -> right = creatTree(pre_begin + index - in_begin + 1, index + 1, in_end, preorder, inorder); return root; }};
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