Comment on page

199.Binary-Tree-Right-Side-View

题目描述

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:
Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:
1 <---
/ \
2 3 <---
\ \
5 4 <---

代码

Approach #1 Two Stacks + DFS

/**
* 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) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
int max_depth = -1;
Stack<TreeNode> nodeStack = new Stack<TreeNode>();
Stack<Integer> depthStack = new Stack<Integer>();
nodeStack.push(root);
depthStack.push(0);
while (!nodeStack.isEmpty()) {
TreeNode node = nodeStack.pop();
int depth = depthStack.pop();
if (node != null) {
max_depth = Math.max(max_depth, depth);
if (!map.containsKey(depth)) {
map.put(depth, node.val);
}
nodeStack.push(node.left);
nodeStack.push(node.right); // 先弹出右节点
depthStack.push(depth + 1);
depthStack.push(depth + 1);
}
}
List<Integer> rightView = new ArrayList<Integer>();
for (int depth = 0; depth <= max_depth; depth++) {
}
return rightView;
}
}

Approach #2 Two Queues + BFS

class Solution {
public List<Integer> rightSideView(TreeNode root) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
int max_depth = -1;
while (!nodeQueue.isEmpty()) {
TreeNode node = nodeQueue.remove();
int depth = depthQueue.remove();
if (node != null) {
max_depth = Math.max(max_depth, depth);
map.put(depth, node.val);