366.寻找二叉树的叶子节点

366.寻找二叉树的叶子节点

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          

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

      1          

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

  []         

Returns [4, 5, 3], [2], [1].

题目大意:将二叉树的叶子节点一层层剪掉,并按层次输出. 阶梯思路:深度优先遍历二叉树(DFS),在每个节点处计算高度(到叶子的距离),这个高度可作为剪枝时的顺序依据,高度越小的越先剪掉. 代码: 特别注意,将vector<vector<int>> res作为递归的参数传入时,必须使用引用&,否则所做修改只在函数作用域内,不会改变外部.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> res;
        dfs(root, res);
        return res;
    }
    //辅助函数,深度遍历
    int dfs(TreeNode* root, vector<vector<int>>& res) {
        if (!root) return 0;  //空节点,高度为0
        int level = max(dfs(root->left, res), dfs(root->right, res)) + 1;  //每个节点高度为子树的最大高度+1,递归求
        if (level > res.size()) res.push_back(vector<int>());  //每当高度超过res长度时,res扩张
        //res从0开始,而level从1开始
        res[level-1].push_back(root->val);  //每个节点写入
        return level;
    }
};