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] ]
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
逐层遍历二叉树，每层的值保存到一个vector。与Binary Tree Level Order Traversal不同的地方在于，本题要求从二叉树的底层开始向上输出。
/**
* 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;
if(root==NULL)return result;
stack<vector<int>> st_result;
queue<TreeNode*>q1;
queue<TreeNode*>q2;
//初始化q1
q1.push(root);
while(!q1.empty() || !q2.empty()){
vector<int>sequence;
if(!q1.empty()){
while(!q1.empty()){
TreeNode*node = q1.front(); q1.pop();
sequence.push_back(node->val);
//将下层节点保存到q2
if(node->left)q2.push(node->left);
if(node->right)q2.push(node->right);
}
}
else{
while(!q2.empty()){
TreeNode*node = q2.front(); q2.pop();
sequence.push_back(node->val);
//将下层节点保存到q2
if(node->left)q1.push(node->left);
if(node->right)q1.push(node->right);
}
}
st_result.push(sequence);
}
//调转输出顺序
while(!st_result.empty()){
result.push_back(st_result.top());
st_result.pop();
}
return result;
}
};LeetCode: Binary Tree Level Order Traversal II [107],布布扣,bubuko.com
LeetCode: Binary Tree Level Order Traversal II [107]
原文：http://blog.csdn.net/harryhuang1990/article/details/28384041