😑366 find leaves of binary tree

https://leetcode.com/problems/find-leaves-of-binary-tree/

先从上向下建立map和indegrees，然后从下向上做拓扑排序

一般来说map记录的是prerequisite -> postrequisite，而indgrees要么是记录node -> indegrees，要么就是postrequisite -> 它的prerequisite，然后通过size看入度，map之所以要这么别扭是因为每次拿到node -> 找postrequisite -> 对每个postrequirsite更新入度 所以是prerequisite -> postrequisite -> postrequisite -> 它的prerequisite的链条

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Map<TreeNode, Integer> indegrees;
    private Map<TreeNode, TreeNode> map;
    private Queue<TreeNode> queue;
    public List<List<Integer>> findLeaves(TreeNode root) {
        indegrees = new HashMap<>();
        map = new HashMap<>();
        queue = new ArrayDeque<>();
        getInfo(root);
        List<List<Integer>> res = new ArrayList<>();
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> ans = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                ans.add(node.val);
                TreeNode par = map.get(node);
                if (par == null) {
                    continue;
                }
                indegrees.put(par, indegrees.get(par)-1);
                if (indegrees.get(par) == 0) {
                    queue.offer(par);
                }
            }
            
            res.add(ans);
        }
        return res;
    }
    private void getInfo(TreeNode node) {
        if (node == null) {
            return;
        }
        indegrees.put(node, indegrees.getOrDefault(node, 0));
        if (node.left != null) {
            indegrees.put(node, indegrees.getOrDefault(node, 0)+1);
            map.put(node.left, node);
            getInfo(node.left);
        }
        if (node.right != null) {
            indegrees.put(node, indegrees.getOrDefault(node, 0)+1);
            map.put(node.right, node);
            getInfo(node.right);
        }
        if (node.left == null && node.right == null) {
            queue.offer(node);
        }
    }
}

当然也可以dfs做，其实就是把高度相同的node放在一起

# 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 findLeaves(self, root: Optional[TreeNode]) -> List[List[int]]:
        self.ans = []
        self.dfs(root)
        return self.ans
    def dfs(self, root):
        if not root:
            return 0
        height = max(self.dfs(root.left), self.dfs(root.right)) + 1
        if height > len(self.ans):
            self.ans.append([])
        self.ans[height - 1].append(root.val)
        return height