Binary Tree Right Side View
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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]
.
解法:
对于树的问题,最常见的就是利用递归,因此这题也利用递归实现。我们从上图可以简单的知道,从右边所看到的点从上到下分别是:顶点,右子树所能看到的点,左子树所能看到的比右子树要深的那些点(忽略右子树,视为透明)。因此,设f(node)返回从右边所看到的点,我们可以得到这样的一个递归式:
①若node为空,则看不到任何点,即返回空
②否则,f(node)=node->val+f(node->right)+f(node->left)[size( f(node->right) ),-1]
class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> temp; if(root==NULL) return temp; else { //顶点先放 temp.push_back(root->val); //右边的点后放 vector<int> right; right=rightSideView(root->right); for(int i=0;i<right.size();++i) temp.push_back(right[i]); //左边的点index大于等于右边个数的点也放 vector<int> left; left=rightSideView(root->left); for(int i=right.size();i<left.size();++i) temp.push_back(left[i]); return temp; } }};
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