[LeetCode] 199. Binary Tree Right Side View 二叉树的右侧视图

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.

Example:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

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

 

给一个二叉树，想象你站在它的右边，返回你能看到的从上到下节点。实际上是二叉树层序遍历的一种变形，只需要保存每一层最右边的数字即可。

解法1：DFS

解法2:  BFS

 

Java: 

public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        rightView(root, result, 0);
        return result;
    }
    
    public void rightView(TreeNode curr, List<Integer> result, int currDepth){
        if(curr == null){
            return;
        }
        if(currDepth == result.size()){
            result.add(curr.val);
        }
        
        rightView(curr.right, result, currDepth + 1);
        rightView(curr.left, result, currDepth + 1);
        
    }
}　　

Python:　DFS　

# Time:  O(n)
# Space: O(h)
class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution(object):
    # @param root, a tree node
    # @return a list of integers
    def rightSideView(self, root):
        result = []
        self.rightSideViewDFS(root, 1, result)
        return result

    def rightSideViewDFS(self, node, depth, result):
        if not node:
            return

        if depth > len(result):
            result.append(node.val)

        self.rightSideViewDFS(node.right, depth+1, result)
        self.rightSideViewDFS(node.left, depth+1, result)

Python: BFS　　

# Time:  O(n)
# Space: O(n)
class Solution2(object):
    # @param root, a tree node
    # @return a list of integers
    def rightSideView(self, root):
        if root is None:
            return []

        result, current = [], [root]
        while current:
            next_level = []
            for node in current:
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)                
            result.append(node.val)
            current = next_level

        return result

Python: Compute the right view of both right and left left subtree, then combine them. For very unbalanced trees, this can be O(n^2), though.

def rightSideView(self, root):
    if not root:
        return []
    right = self.rightSideView(root.right)
    left = self.rightSideView(root.left)
    return [root.val] + right + left[len(right):]

Python: DFS-traverse the tree right-to-left, add values to the view whenever we first reach a new record depth. This is O(n).

def rightSideView(self, root):
    def collect(node, depth):
        if node:
            if depth == len(view):
                view.append(node.val)
            collect(node.right, depth+1)
            collect(node.left, depth+1)
    view = []
    collect(root, 0)
    return view　

Python: Traverse the tree level by level and add the last value of each level to the view. This is O(n).

def rightSideView(self, root):
    view = []
    if root:
        level = [root]
        while level:
            view += level[-1].val,
            level = [kid for node in level for kid in (node.left, node.right) if kid]
    return view　　

C++: DFS

class Solution {
public:
    void recursion(TreeNode *root, int level, vector<int> &res)
    {
        if(root==NULL) return ;
        if(res.size()<level) res.push_back(root->val);
        recursion(root->right, level+1, res);
        recursion(root->left, level+1, res);
    }
    
    vector<int> rightSideView(TreeNode *root) {
        vector<int> res;
        recursion(root, 1, res);
        return res;
    }
};

C++:　BFS　

class Solution {
public:
    vector<int> rightSideView(TreeNode *root) {
        vector<int> res;
        if (!root) return res;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            res.push_back(q.back()->val);
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                TreeNode *node = q.front();
                q.pop();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return res;
    }
};

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[LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历

[LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 II

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