LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

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.

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

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题目大意:

给定满二叉树中序遍历和先序遍历，建立二叉树。
可以根据先序遍历找出根节点，在中序遍历中找出该节点元素，该节点元素的左边都是左子树的节点，右边都是右子树的节点。可以通过递归建树。

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题目代码:

/** * 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 build(preorder, inorder, 0, preorder.size()-1, 0, inorder.size()-1);    }        TreeNode* build(vector<int>& preorder, vector<int>& inorder, int ps, int pe, int is, int ie){        if(ps > pe || is > ie)            return nullptr;        TreeNode* root = new TreeNode(preorder[ps]);        int i = 0;        while(inorder[i] != root->val) i++;        root->left  = build(preorder, inorder, ps+1, ps+i-is, is, i-1);        root->right = build(preorder, inorder, ps+i-is+1, pe,i+1, ie);        return root;    }    };