leetcode-199-Binary-Tree-Right-Side-View

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描述

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:

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

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

分析

广度优先遍历，拿 515. Find Largest Value in Each Tree Row 的代码改了两行就 AC 了。

解决方案1（Python）

深度优先搜索，因为每次压栈然后出栈，会先访问右子树，对于每一层来说，见到的都是最右边的节点。

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        depthValueMap = dict()
        maxDepth = -1
        
        stack = [(root, 0)]
        
        while stack:
            node, depth = stack.pop()
            
            if node is not None:
                maxDepth = max(maxDepth, depth)
                depthValueMap.setdefault(depth, node.val)
                stack.append((node.left, depth+1))
                stack.append((node.right, depth+1))
        return [depthValueMap[depth] for depth in range(maxDepth+1)]

解决方案1(Java)

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new LinkedList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int nowRightNode = Integer.MIN_VALUE, size = queue.size();
        while(!queue.isEmpty()) {
            size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode nowNode = queue.poll();
                if (i == size-1) {
                    nowRightNode = nowNode.val;
                }
                if (nowNode.left != null) {
                    queue.offer(nowNode.left);
                }
                if (nowNode.right != null) {
                    queue.offer(nowNode.right);
                }
            }
            result.add(nowRightNode);
        }
        return result;
    }
}

解决方案2（golang）

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func rightSideView(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }
    result := []int{}
    queue := make([]*TreeNode, 0)
    queue = append(queue, root)
    
    for len(queue) > 0 {
        length := len(queue)
        result = append(result, queue[length-1].Val)
        for i := 0; i < length; i++ {
            node := queue[0]
            queue = queue[1:]
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
    }
    return result
}

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