LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

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.

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。

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;
    }
    
};