leetcode366- Find Leaves of Binary Tree- medium

Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

Example:
Given binary tree 

          1
         / \
        2   3
       / \     
      4   5    

Returns [4, 5, 3], [2], [1].

Explanation:

1. Removing the leaves [4, 5, 3] would result in this tree:

          1
         / 
        2          

2. Now removing the leaf [2] would result in this tree:

          1          

3. Now removing the leaf [1] would result in the empty tree:

          []         

Returns [4, 5, 3], [2], [1].

 

算法：DFS递归，边计算节点深度边做。helper里做的是把深度相同的节点放到一个map的记录里。之后主函数就可以直接对深度相同的一类节点的值放到一个list。

数据结构：Map<Integer, Set<TreeNode>>。记录某个深度(integer)对应的所有树节点(Set<TreeNode>)

细节：1.map.containsKey()不要只写contains()；map.keySet()，k不要大写。  2.List<Integer>里是.add(node.val)不是add(node)  3.每次private函数名字、参数改了一定要到所有调用这个函数的地方都改一遍！尤其是递归的时候千万小心！

 

实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> findLeaves(TreeNode root) {
        
        List<List<Integer>> result = new ArrayList<>();
        Map<Integer, Set<TreeNode>> map = new HashMap<>();
        helper(root, map);
        for (int i = 1; i <= map.keySet().size(); i++) {
            List<Integer> l = new ArrayList<>();
            for (TreeNode n : map.get(i)) {
                l.add(n.val);
            }
            result.add(l);
        }           
        return result;
    }
    
    private int helper(TreeNode root, Map<Integer, Set<TreeNode>> map) {
        
        if (root == null) {
            return 0;
        }
        
        int left = helper(root.left, map);
        int right = helper(root.right, map);
        int depth = Math.max(left, right) + 1;
        
        if (!map.containsKey(depth)) {
            map.put(depth, new HashSet<TreeNode>());
        }
        map.get(depth).add(root);
        
        return depth;
    }
}