【LeetCode】105. Construct Binary Tree from Preorder and Inorder Traversal
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题目描述:
Given preorder and inorder traversal of a tree, construct the binary tree.
给定一棵二叉树的先序遍历与中序遍历,构造出这棵二叉树。
解题思路:
对于一棵二叉树:记某个节点为V, 其左子树为L,右子树为R
先序遍历:VLR
中序遍历:LVR
后序遍历:LRV
得到二叉树的先序遍历VLR与中序遍历LVR后,
先序遍历的第一个元素为V,即父节点。
得到父节点V后,从中序遍历中找到V的位置,从而分别得到左子树与右子树的元素个数。
再利用递归求出左子树与右子树的构成。
值得注意的是,递归的退出条件为元素个数为0.
要注意递归参数的取值。
参考代码:
/** * 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) { // preorder VLR // inorder LVR int i = 0, j = preorder.size(); int m = 0, n = inorder.size(); return buildT(preorder,i,j,inorder,m,n); } TreeNode* buildT(vector<int>& preorder, int i, int j, vector<int>& inorder, int m, int n){ //empty if (j - i <= 0)return 0; TreeNode *node = new TreeNode(preorder[i]); int k = 0; for(;k<inorder.size();++k){ if (inorder[k] == preorder[i])break; } int left = k - m; int right = n - k; node->left = buildT(preorder,i+1,i+1+left,inorder,m,m+left); node->right = buildT(preorder,i+1+left+1,j,inorder,m+left+1,n); return node; }};
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