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LeetCode--construct-binary-tree-from-preorder-and-inorder-traversal

2018-01-07 20:36 537 查看


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.

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *dfs(vector<int> &preorder,int preStart,int preEnd,vector<int> &inorder,int inStart,int inEnd)
{
    if(preStart > preEnd)
        return nullptr;
    TreeNode *root = new TreeNode(preorder[preStart]);
    int middle;
    for(middle=inStart;middle<=inEnd;middle++)
        if(inorder[middle] == root->val)
            break;
    int leftLen = middle-inStart;
    root->left = dfs(preorder,preStart+1,preStart+leftLen,inorder,inStart,middle-1);
    root->right = dfs(preorder,preStart+leftLen+1,preEnd,inorder,middle+1,inEnd);
    return root;
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder)
{
    int pre = preorder.size(),in = inorder.size();
    if(pre==0 || in==0 || pre != in)
        return nullptr;
    return dfs(preorder,0,pre-1,inorder,0,in-1);
}
};
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