LeetCode Binary Tree Level Order Traversal II

最新推荐文章于 2021-11-18 16:57:58 发布

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最新推荐文章于 2021-11-18 16:57:58 发布

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分类专栏：算法设计与分析 OJ

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本文介绍了一种解决二叉树层次遍历问题的方法，即从叶子节点开始向上逐层遍历并返回各层节点值。通过使用广度优先搜索（BFS），先按从下到上的顺序收集节点，再反转结果以实现从上到下的层次输出。

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Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

题意：给出一个二叉树，按层次从高到低输出

思路：用BFS，先按层次从低到高，然后倒过来

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        queue<TreeNode*> q;
	vector<vector<int> > vv;
	vector<int> v;
	
	if (root) {
		v.push_back(root->val);
		vv.push_back(v);
	}
	
	q.push(root);
	vector<TreeNode*> vt;
	while (!q.empty()) {
		TreeNode* cur = q.front(); q.pop();
		vt.push_back(cur);
		if (cur == NULL) continue;
		q.push(cur->left);
		q.push(cur->right);
	}
	
	int step = 2;
	size_t j;
	
	for (size_t i = 1; i < vt.size(); i = j) {
		v.clear();
		
		int cnt = 0;
		
		for (j = i; j < i + step && j < vt.size(); j++) {
			if (vt[j]) {
				v.push_back(vt[j]->val);
				cnt += 2;
			}
		}
		
		step = cnt;
		if (v.size() > 0) {
			vv.push_back(v);
		}
	}
	
	reverse(vv.begin(), vv.end());
	return vv;
    }
};