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[LeetCode] Binary Tree Right Side View

来源：程序员人生发布时间：2015-06-05 09:07:29 阅读次数：2084次

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].

解题思路

层次遍历法，找出每层最右真个结点。

实现代码1

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

 // Runtime:7ms
class Solution {
public:
    vector<int> rightSideView(TreeNode *root) {
        vector<int> nums;
        queue<TreeNode*> nodes;
        if (root != NULL)
        {
            nodes.push(root);
        }

        TreeNode *cur;
        while (!nodes.empty())
        {
            int size = nodes.size();
            for (int i = 0; i < size; i++)
            {
                cur = nodes.front();
                nodes.pop();
                if (cur->left != NULL)
                {
                    nodes.push(cur->left);
                }
                if (cur->right != NULL)
                {
                    nodes.push(cur->right);
                }
            }

            nums.push_back(cur->val);
        }

        return nums;
    }
};

实现代码2

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# Runtime:74ms
class Solution:
    # @param root, a tree node
    # @return a list of integers
    def rightSideView(self, root):
        nums = []
        if root == None:
            return nums
        nodes = [root]

        while nodes:
            size = len(nodes)
            for i in range(size):
                cur = nodes.pop(0)
                if cur.left != None:
                    nodes.append(cur.left)
                if cur.right != None:
                    nodes.append(cur.right)
            nums.append(cur.val)

        return nums;

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