LeetCode 107. Binary Tree Level Order Traversal II

题目描述:

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]
]

先层次遍历, 然后将得到的二维数组颠倒顺序.

/**
 * 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 {
    vector<vector<int>> ret;
public:
	vector<vector<int>> levelOrderBottom(TreeNode* root) {
        if(!root) return ret;
        BFS(root);
        reverse(ret.begin(), ret.end());
		return ret;
	}
	
	void BFS(TreeNode *root){
	    queue<TreeNode*> nodeQueue;
	    queue<int> levelQueue;
	    
	    nodeQueue.push(root);
	    levelQueue.push(0);
	    
	    while(!nodeQueue.empty()){
	        TreeNode *node = nodeQueue.front();
	        int nodeLevel = levelQueue.front();
	        nodeQueue.pop();
	        levelQueue.pop();
	        
	        if(nodeLevel >= ret.size()){
	            ret.resize(nodeLevel + 1);
	        }
	        ret[nodeLevel].push_back(node->val);
	        
	        if(node->left){
	            nodeQueue.push(node->left);
	            levelQueue.push(nodeLevel + 1);
	        }
	        if(node->right){
	            nodeQueue.push(node->right);
	            levelQueue.push(nodeLevel + 1);
	        }
	    }
	}
};