Construct Binary Tree from Preorder and Inorder Traversal

題目

leetcode 105， https://leetcode.com/problems...

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

3

/ \
9 20

/  \

15 7

思路

前序遍歷的第一個節點就是root，用這個root能夠把中序遍歷的列表正好分爲兩段，分別是左右子樹。左右子樹的size，正好是前序有序的兩段。
所以，能夠根據前序第一個節點，劃分出兩個前序+中序的序列，遞歸生成子樹。再將子樹鏈接到root上。

代碼

/**
 * 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) {
        int size = preorder.size();
        if (size == 0) { return NULL; }
        int p_idx = 0;
        int root_val = preorder[p_idx++];
        TreeNode* root = new TreeNode(root_val);
        vector<int> p1,p2;
        vector<int> i1,i2;
        
        bool in_left = true;
        for (int i = 0; i < size; i++) {
            if (inorder[i] == root_val) {
                in_left = false;
                continue;
            }
            if (in_left) {
                i1.push_back(inorder[i]);
                p1.push_back(preorder[p_idx++]);
            } else {
                i2.push_back(inorder[i]);
                p2.push_back(preorder[p_idx++]);
            }
        }
        root->left = buildTree(p1, i1);
        root->right = buildTree(p2, i2);
        return root;
    }
};