leetcode 解题: Construct Binary Tree from Preorder and Inorder Traversal (105)

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

解法1： Recursive

Preorder的顺序是root，left，right
inorder的顺序是left，root，right
可见preorder的第一个一定是root，这样有了root的数值，我们就可以在inorder中找到root的位置，左边都是left，右边都是right
对左右子树同时进行tree construction。要注意的是下标容易出错， 要注意。

C++

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/**
 * 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() != inorder.size()) {
            return NULL;
        }
        return helper(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
    }
    TreeNode* helper(vector<int>& preorder, int pstart, int pend, vector<int>& inorder, int istart, int iend) {
        if (pstart > pend || istart > iend) {
            return NULL;
        }
        if (pstart == pend) {
            return new TreeNode(preorder[pstart]);
        }
        int rootval = preorder[pstart];
        TreeNode* root = new TreeNode(rootval);
        auto iter = find(inorder.begin() + istart, inorder.begin() + iend + 1, rootval);
        int leftNum = iter - (inorder.begin() + istart);
        TreeNode* left = helper(preorder, pstart + 1, pstart + leftNum, inorder, istart, istart + leftNum - 1);
        TreeNode* right = helper(preorder, pstart + leftNum + 1, pend, inorder, istart + leftNum + 1, iend);
        root->left = left;
        root->right = right;
        return root;
    }
};

Java

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class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return helper(0, 0, inorder.length - 1, preorder, inorder);
    }
    public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
        if (preStart > preorder.length - 1 || inStart > inEnd) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int inIndex = 0; // Index of current root in inorder
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == root.val) {
                inIndex = i;
            }
        }
        root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
        root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);
        return root;
    }
}