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Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
分层遍历二叉树，用BFS即可
class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { queue<TreeNode*> Q; vector<vector<int>> ans; Q.push(root); while(!Q.empty()) { int e = Q.size(); //当前层的结束位置 vector<int> partans; for(int i = 0; i < e; i++) { TreeNode * p = Q.front(); Q.pop(); if(NULL != p) { partans.push_back(p->val); Q.push(p->left); Q.push(p->right); } } if(!partans.empty()) ans.push_back(partans); } return ans; } };
网上DFS的代码：
class Solution { protected: vector<vector<int>> ans; void dfs(TreeNode *root, int height){ if (root == NULL) return; while (ans.size() <= height) ans.push_back(vector<int>()); ans[height].push_back(root->val); dfs(root->left, height + 1); dfs(root->right, height + 1); } public: vector<vector<int>> levelOrder(TreeNode* root) { dfs(root, 0); return ans; } };
Binary Tree Level Order Traversal II
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},
3
/ 9 20
/ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
跟上面只是顺序变了，加个reverse就行了。
【leetcode】Binary Tree Level Order Traversal I & II
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原文地址：http://www.cnblogs.com/dplearning/p/4627247.html
