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[Google] LeetCode 366 Find Leaves of Binary Tree 思维+DFS

作者：互联网

Given the root of a binary tree, collect a tree's nodes as if you were doing this:

Solution

给定一个二叉树，每次输出叶子节点，然后去除叶子节点，不断重复直到根节点也被删除。这道题的思路其实感性来看就是要剥开每一层的节点，然后存储起来。

实际上我们也不需要真正将节点去除。我们要记录每一个节点到叶子节点的距离，这里记为 \(lev\)，叶子节点到本身的距离记为 \(1\)，我们通过 \(DFS\) 来得到：

点击查看代码 
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int dfs(vector<vector<int>>& ans, TreeNode* root){
        if(root==NULL)return 0;
        int lev = max(dfs(ans, root->left), dfs(ans, root->right))+1;
        if(lev>ans.size())ans.push_back(vector<int>());
        ans[lev-1].push_back(root->val);
        return lev;
    }
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> ans;
        dfs(ans, root);
        return ans;
    }
};

标签：Binary,Google,TreeNode,Leaves,right,ans,root,节点,left
来源： https://www.cnblogs.com/xinyu04/p/16492607.html