#Leetcode# 107. Binary Tree Level Order Traversal II

https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

 

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

代码：

/**
 * 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>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> > ans;
        if(!root) return ans;
        level(root, ans, 1);
        for(int i = 0; i < ans.size() / 2; i ++)
            swap(ans[i], ans[ans.size() - 1 - i]);
        return ans;
    }
    void level(TreeNode* root, vector<vector<int> >& ans, int depth) {
        vector<int> v;
        if(depth > ans.size()) {
            ans.push_back(v);
            v.clear();
        }
        ans[depth - 1].push_back(root -> val);
        if(root -> left)
            level(root -> left, ans, depth + 1);
        if(root -> right)
            level(root -> right, ans, depth + 1);
    }
};

　　和之前的一个差不多 层序遍历然后把数组反过来可以了 是 Tree 的第 $20$ 题