Binary Tree Level Order Traversal II
2014-12-14 19:34
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https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/
1、题目要求:给定一个二叉树,要求从底层到顶层层序遍历二叉树(从叶结点到根结点,从左至右遍历二叉树)
例:
给定二叉树{3, 9, 20, #, #, 15, 7}
其从底层到顶层遍历的结果是:
2、解题思路:首先用两个队列que来实现分层遍历,然后将每一层的结点入栈,即用栈来实现对层序遍历的逆序。
注:此时栈stk中保存的不是单个结点,而是一层的所有结点。
1、题目要求:给定一个二叉树,要求从底层到顶层层序遍历二叉树(从叶结点到根结点,从左至右遍历二叉树)
例:
给定二叉树{3, 9, 20, #, #, 15, 7}
3 / \ 9 20 / \ 15 7
其从底层到顶层遍历的结果是:
[ [15,7], [9,20], [3] ]
2、解题思路:首先用两个队列que来实现分层遍历,然后将每一层的结点入栈,即用栈来实现对层序遍历的逆序。
注:此时栈stk中保存的不是单个结点,而是一层的所有结点。
/** * 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) { vector<vector<int> > result; vector<int> vec; stack<vector<int> > stk; //用栈stk来实现层次遍历的逆序 queue<TreeNode *> que1, que2; if(!root) return result; else { que1.push(root); while(!que1.empty()||!que2.empty()) //二叉树的层序遍历 { while(!que1.empty()) { if(que1.front()->left) que2.push(que1.front()->left); if(que1.front()->right) que2.push(que1.front()->right); vec.push_back(que1.front()->val); que1.pop(); } if(!vec.empty()) stk.push(vec); //将每一层的结点数据入栈stk; vec.clear(); while(!que2.empty()) { if(que2.front()->left) que1.push(que2.front()->left); if(que2.front()->right) que1.push(que2.front()->right); vec.push_back(que2.front()->val); que2.pop(); } if(!vec.empty()) stk.push(vec); //将每一层的结点数据入栈stk; vec.clear(); } while(!stk.empty()) //层序遍历的逆序 { result.push_back(stk.top()); stk.pop(); } return result; } } };
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