LeetCode:Binary Tree Right Side View
2015-04-04 22:50
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原题:
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,
You should return
题意:从右侧看二叉树,所看到的每一层最右侧的节点,该节点可以在右子树上,可以在左子树上。
解题思路:层序遍历,在每一层从右往左访问节点,这样在每下降一层的时候都会首先访问最右边的节点,这个节点就是我们想要的节点,我们用markedLevel表示已经标示的最大层,即已经选出最右节点的层,用currentLevel标示正在访问的节点所在的层,这样,当第一次currentLevel>
markedLevel的时候,表示我们正在访问最右节点。采用递归算法,思路清晰,代码量少。Java实现如下:
参考地址:http://www.geeksforgeeks.org/print-right-view-binary-tree-2/
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].
题意:从右侧看二叉树,所看到的每一层最右侧的节点,该节点可以在右子树上,可以在左子树上。
解题思路:层序遍历,在每一层从右往左访问节点,这样在每下降一层的时候都会首先访问最右边的节点,这个节点就是我们想要的节点,我们用markedLevel表示已经标示的最大层,即已经选出最右节点的层,用currentLevel标示正在访问的节点所在的层,这样,当第一次currentLevel>
markedLevel的时候,表示我们正在访问最右节点。采用递归算法,思路清晰,代码量少。Java实现如下:
public class Solution { ArrayList<Integer> list = new ArrayList<Integer>(); int markedLevel = 0; public void levelDown(TreeNode root,int currentLevel){ if(root == null){ return; } if(currentLevel>markedLevel){ //第一次currentLevel大于markedLevel表示访问最右节点 list.add(root.val); markedLevel = currentLevel; //不再记录当前层 } levelDown(root.right,currentLevel+1); //总是先访问每一层的最右节点 levelDown(root.left,currentLevel+1); } public List<Integer> rightSideView(TreeNode root) { levelDown(root,1); return list; } }变通:如果需要输出的是二叉树的最左视图,只需要每次首先访问左侧节点即可。
参考地址:http://www.geeksforgeeks.org/print-right-view-binary-tree-2/
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