leetcode-14-basic-breadthFirstSearch

BFS: breadth first search



107. Binary Tree Level Order Traversal II

解题思路:

本来我是用map<int,int>存所有节点的值和深度(root是0),然后遍历map,result[depth].push_back(val)。但是因为map是无序的,所以

插入的时候,result[i]里元素的顺序会有问题,比如

后面改用下面的方法。先计算树的最大深度,然后遍历树的时候直接插入。后面自己写了个测试,有点丑(微笑脸)。

#include <iostream>
#include <cmath>
#include <vector>
#include <ctime>
#include <time.h>
#include <stdlib.h>
#include <map>
#include <algorithm>
using namespace std;

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

class Solution {
public:
	// if leaf node, depth=1
	int Depth(TreeNode* root) {
		if (!root)
		    return 0;
		return max(Depth(root->left), Depth(root->right)) + 1;
	}
	void make(TreeNode* root, int depth, vector<vector<int> >& result) {
		if (!root)
		    return;
		// insert
		result[depth].push_back(root->val);
		make(root->left, depth-1, result);
		make(root->right, depth-1, result);
	}
    vector<vector<int> > levelOrderBottom(TreeNode* root) {
    	int depth = Depth(root);
    	// result needs initialization
    	vector<vector<int> > result(depth, vector<int> {});
    	make(root, depth-1, result);
    	return result;
	}

};

int main() {
	Solution s;
	TreeNode temp(3);
	TreeNode* root = &temp;
	TreeNode a(9);
	root->left = &a;
	TreeNode b(20);
	root->right = &b;
	TreeNode c(15);
	root->right->left = &c;
	TreeNode d(7);
	root->right->right = &d;
	vector<vector<int> > re = s.levelOrderBottom(root);
//	cout << re.empty();
	vector<vector<int> >::iterator it1;
	vector<int>::iterator it2;
	for (it1 = re.begin(); it1 != re.end(); it1++) {
		cout << "*" << " ";
		for (it2 = it1->begin(); it2 != it1->end(); it2++)
		    cout << *it2 << " ";
		cout << endl;
	}
	return 0;
}

类似的题目有:

102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).

解题思路:

这个是要正序输出,那么只要改动make中

make(root->left, depth+1, result);
make(root->right, depth+1, result);

改动levelOrder中

make(root, 0, result);

即可。

类似的还有:

515. Find Largest Value in Each Tree Row

解题思路:可以用上面的方法找每行的元素,然后取最大值就可以了。不过比较慢。

改进了一下,只存最大值就好了。需要注意的是,测试用例中有负数。。不只是int型,所以用一个超小值做初始值。

vector<int> largestValues(TreeNode* root) {
        vector<int> result2;
        if (!root)
            return result2;
        int depth = Depth(root);
        int size = depth;
    	// result needs initialization
    	vector<int> result(depth, -2147483648);
    	make(root, 0, result);
    	return result;
    }
	// if leaf node, depth=1
	int Depth(TreeNode* root) {
		if (!root)
		    return 0;
		return max(Depth(root->left), Depth(root->right)) + 1;
	}
	void make(TreeNode* root, int depth, vector<int>& result) {
		if (!root)
		    return;
		if (root->val > result[depth])
		    result[depth] = root->val;
		make(root->left, depth+1, result);
		make(root->right, depth+1, result);
	}

好吧,类似的还有这道:

513. Find Bottom Left Tree Value

解题思路:

仍然是复用上面的代码,改

return result[result.size()-1][0];

不过显然这样效率很低。我想的改进是,增加一个记录树深度的变量d,在make函数中,push_back后增加一个判断,

如果已经到了最后一层,就终止,不再压后面的栈。这样的话,时间从22s->12s。

if (depth == d-1)
    return;


leetcode-14-basic-breadthFirstSearch

时间: 2024-10-07 07:46:25

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