LeetCode——Construct Binary Tree from Preorder and Inorder Traversal

Question

Given preorder and inorder traversal of a tree, construct the binary tree.

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

Solution

参考：http://www.cnblogs.com/zhonghuasong/p/7096150.html

Code

/**
 * 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) {
        if (preorder.size() == 0 || inorder.size() == 0)
            return NULL;
            
        return ConstructTree(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
    }
    
    TreeNode* ConstructTree(vector<int>& preorder, vector<int>& inorder, 
            int pre_start, int pre_end, int in_start, int in_end) {
        int rootValue = preorder[pre_start];
        TreeNode* root = new TreeNode(rootValue);
        if (pre_start == pre_end) {
            if (in_start == in_end)
                return root;
        }
        
        int rootIn = in_start;
        while (rootIn <= in_end && inorder[rootIn] != rootValue)
            rootIn++;
            
        int preLeftLength = rootIn - in_start;
        if (preLeftLength > 0) {
            root->left = ConstructTree(preorder, inorder, pre_start + 1, preLeftLength, in_start, rootIn - 1);
        }
        // 左子树的对大长度就是in_end - in_start
        if (preLeftLength < in_end - in_start) {
            root->right = ConstructTree(preorder, inorder, pre_start + 1 + preLeftLength, pre_end, rootIn + 1, in_end);
        }
        
        return root;
    }
};