标签：
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,null,null,15,7],
3
/ 9 20
/ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
BST反着输出
类似Binary Tree Level Order Traversal I，用栈实现，最后逆序即可
/**
* Definition for a binary tree node.
* 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>> res;
if(root==NULL)return res;
vector<int> v;
queue<TreeNode*> q;
q.push(root);
int curCount=1;
int nextCount=0;
while(!q.empty())
{
TreeNode* curNode=q.front();
q.pop();
v.push_back(curNode->val);
curCount--;
if(curNode->left)
{
nextCount++;
q.push(curNode->left);
}
if(curNode->right)
{
nextCount++;
q.push(curNode->right);
}
if(curCount==0)
{
curCount=nextCount;
nextCount=0;
res.push_back(v);
v.clear();
}
}
reverse(res.begin(),res.end());
return res;
}
};
leetcode No107. Binary Tree Level Order Traversal II
标签：
原文地址：http://blog.csdn.net/u011391629/article/details/52274042
