leetCode(21):Binary Tree Right Side View

标签：leetcode   二叉树   

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /   2     3         <---
 \       5     4       <---

You should return [1, 3, 4].

采用层序遍历的方式，每一层的最后一个结点即为在右侧可以看到的结点；

/**
 * 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<int> rightSideView(TreeNode* root) {
        queue<TreeNode*> nodes;
    	vector<int> vec;
    	vector< vector<int> > result;
    	vector<int> tmp;
    	if(NULL==root)
    		return vec;
    	nodes.push(root);
    	while(!nodes.empty())
    	{
    		int length=nodes.size();
    		int i=0;
    		while(i<length)
    		{
    			TreeNode* tmpNode=nodes.front();
    			tmp.push_back(tmpNode->val);
    			if(tmpNode->left)
    				nodes.push(tmpNode->left);
    			if(tmpNode->right)
    				nodes.push(tmpNode->right);
    			nodes.pop();
    			i++;
    		}
    		result.push_back(tmp);
    		tmp.clear();
    	}
    	
    	for(size_t i=0;i<result.size();++i)
    	{
    		vec.push_back(result[i][result[i].size()-1]);
    	}
    	return vec;
    }
};

leetCode(21):Binary Tree Right Side View

标签：leetcode   二叉树   

原文地址：http://blog.csdn.net/walker19900515/article/details/46591313