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,#,#,15,7},

3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

跟从上往下层次遍历一样。最后把结果倒置一下就能够了~~

struct node {	 TreeNode* tn;	 int level; };		class Solution 	{	public:		vector
     
      
        > levelOrderBottom(TreeNode *root) 		{			vector
       
        
          > vvi;			vector
         
           vi;			deque
          
            di; node nd; int level = 0; if (root == NULL) { return vvi; } nd.level = 0; nd.tn = root; di.push_back(nd); node left,right; while (!di.empty()) { nd = di.front(); if (nd.tn->left != NULL) { left.tn = nd.tn->left; left.level = nd.level+1; di.push_back(left); } if (nd.tn->right != NULL) { right.tn = nd.tn->right; right.level = nd.level + 1; di.push_back(right); } // if (vi.empty()) // { // vi.push_back(nd.tn->val); // level++; // di.pop_front(); // } // else // { nd = di.front(); if (nd.level == level) { vi.push_back(nd.tn->val); di.pop_front(); } else { vvi.push_back(vi); vi.clear(); vi.push_back(nd.tn->val); level++; di.pop_front(); } // } } vvi.push_back(vi); return vector
           
            
              >(vvi.rbegin(),vvi.rend()); } };