Leetcode: Construct Binary Tree from Preorder and Inorder Traversal, Construct Binary Tree from Inorder and Postorder Traversal

总结:

1. 第 36 行代码, 最好是按照 len 来遍历, 而不是下标

代码: 前序中序

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 };

class Solution {
public:
	vector<int> preorder, inorder;
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
		TreeNode * root = NULL;
		if(preorder.size() == 0 || inorder.size() == 0)
			return root;

		this->preorder = preorder;
		this->inorder = inorder;
		for(int i = 0; i < inorder.size(); i ++) {
			if(inorder[i] == preorder[0]) {
				root = new TreeNode(preorder[0]);
				int len1 = i;
				int len2 = inorder.size()-i-1;
				root->left = buildParty(1,0, len1);
				root->right = buildParty(len1+1, i+1, len2);
				return root;
			}
		}
    }
	TreeNode *buildParty(const int &p, const int &i, const int &len) {
		if(len <= 0)
			return NULL;
		for(int cursor = 0; cursor < len; cursor++) {
			int pos = cursor+i;

			if(inorder[pos] == preorder[p]) {
				TreeNode *root = new TreeNode(preorder[p]);
				int len1 = cursor;
				int len2 = len-cursor-1;
				root->left = buildParty(p+1, i, len1);
				root->right = buildParty(p+len1+1, pos+1, len2);
				return root;
			}
		}
	}
};

int main() {
	TreeNode *node;

	int in1[10] = {1, 2, 3, 4, 5, 6};
	int in2[10] = {3, 2, 4, 1, 5, 6};

	Solution solution;
	node = solution.buildTree(vector<int>(in1, in1+6), vector<int>(in2, in2+6));
	return 0;
}

　　

代码: 中序后序

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 };

class Solution {
public:
	vector<int> inorder;
	vector<int> postorder;
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
		TreeNode *root = NULL;
		if(!inorder.size())
			return root;

		this->inorder = inorder;
		this->postorder = postorder;

		for(int ci = 0; ci < inorder.size(); ci++) {
			if(inorder[ci] == postorder[postorder.size()-1]) {
				root = new TreeNode(inorder[ci]);
				int len1 = ci;
				int len2 = inorder.size()-ci-1;
				root->left = buildParty(0, postorder.size()-len2-2, len1);
				root->right = buildParty(ci+1, postorder.size()-2, len2);
				return root;
			}

		}
    }
	TreeNode *buildParty(const int &i, const int &j, const int &len) {
		if(!len)
			return NULL;

		for(int ci = 0; ci < len; ci ++) {
			int pos = i+ci;
			if(postorder[j] == inorder[pos]) {
				TreeNode *root = new TreeNode(inorder[pos]);
				int len1 = ci;
				int len2 = len-ci-1;
				root->left = buildParty(i, j-len2-1, len1);
				root->right = buildParty(i+ci+1, j-1, len2);
				return root;
			}
		}
	}
};

int main() {
	TreeNode *node;

	int in1[10] = {3, 2, 4, 1, 5, 6};
	int in2[10] = {3, 4, 2, 6, 5, 1};

	Solution solution;
	node = solution.buildTree(vector<int>(in1, in1+6), vector<int>(in2, in2+6));
	return 0;
}