107 Binary Tree Level Order Traversal II

1 题目

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]
]

2 尝试解

2.1 分析

逆序输出层序遍历结果。

2.2 代码

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        if(root == NULL) return result;
        queue<TreeNode*> layer;
        layer.push(root);
        while(!layer.empty()){
            vector<int> saver;
            int size = layer.size();
            for(int i = 0; i <size; i++){
                TreeNode* cur = layer.front();
                layer.pop();
                saver.push_back(cur->val);
                if(cur->left != NULL) layer.push(cur->left);
                if(cur->right != NULL) layer.push(cur->right);
            }
            result.insert(result.begin(),saver);
        }
        return result;
    }
};

3 标准解

3.1 代码

vector<vector<int> > levelOrder(TreeNode *root) {
	vector<vector<int> > retVal;

	levelOrder(root, retVal, 0);

	reverse(retVal.begin(), retVal.end());

	return retVal;
}

void levelOrder(TreeNode* root, vector<vector<int> > &v, int currLevel) {
	if (root == NULL) {
		return;
	}

	if (v.empty() || currLevel > (v.size() - 1)) {
		v.push_back(vector<int>());
	}

	v[currLevel].push_back(root->val);

	levelOrder(root->left, v, currLevel + 1);
	levelOrder(root->right, v, currLevel + 1);
}