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,#,#,15,7},3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7] [9,20], [3],]
解题思路:
1)我的思路:(操作过于复杂)
先将二叉树镜像 3 / \ 9 20 / \ 15 7从下到上层序: 15 7 | 9 20 | 3 正常的层序:3 | 3 20 | 15 7 3 / \ 20 9 / \7 15从上到下: 3 | 20 9 | 7 15 与原树从下到上层序(要求解的结果)正好是倒叙,可以将结果存入栈中。使用另一个栈记录每层的节点数。从两个栈中,读取最后结果
class Solution {public: vector> levelOrderBottom(TreeNode *root) { vector > result; if(root==nullptr) return result; //将二叉树镜像 MirrorBinaryTree(root); //vector > res; //return res; //层序遍历从左到右,从上到下,存入栈中 stack treeData,numLevel; queue pNode; pNode.push(root); TreeNode * curr; int currCount=1,nextCount=0; numLevel.push(currCount); //存第一个root节点,该变量用于存储每层的节点数 while(!pNode.empty()){ //队列不为空的时候 不能直接写 while(pNode) curr = pNode.front(); pNode.pop(); treeData.push(curr->val); //存入当前节点值 currCount--; if(curr->left){ pNode.push(curr->left); nextCount++; } if(curr->right){ pNode.push(curr->right); nextCount++; } if(currCount==0){ if(nextCount!=0) numLevel.push(nextCount); currCount = nextCount; nextCount = 0; } } //从栈中读取结果 vector row; //int index = 0; while(!numLevel.empty()){ //error:while(numLevel) int num = numLevel.top(); numLevel.pop(); while(num){ num--; row.push_back(treeData.top()); treeData.pop(); } //index++; result.push_back(row); vector ().swap(row);//清空row中的值 } return result; } void MirrorBinaryTree(TreeNode *root){ //返回值为空,在原树上修改,不要增加新的空间。 //终止条件 if(root==nullptr) return ; if((root->left==nullptr) && (root->right==nullptr)) return ; //交换左右子树 TreeNode *temp = root->left; root->left =root->right; root->right = temp; if(root->left!=nullptr) //递归交换左子树 MirrorBinaryTree(root->left); if(root->right!=nullptr) //递归交换右子树 MirrorBinaryTree(root->right); }};
2)广度优先遍历,然后对结果二维数组result的第一个维度做逆转
reverse(res.begin(),res.end()); //逆转的复杂度是不是很大?
class Solution {public: vector> levelOrderBottom(TreeNode *root) { vector > res; if(root==nullptr) return res; queue q; q.push(root); while(q.size()>0) { vector level; for(int i=0,n=q.size();i val); if(temp->left) q.push(temp->left); if(temp->right) q.push(temp->right); } res.push_back(level); } reverse(res.begin(),res.end()); //c++ 使用自带函数 return res; }};
3) DFS 深度优先遍历
思路很简便:初始化二维数组,存取数字时,从二维数组的第一维度的最大值存储(即从最后一行开始存,然后存倒数第二行)
初始化时,要知道二维数组一共有多少行,求树的高度即可。
class Solution {public: int getHeight(TreeNode *root) { if(!root) return 0; return max(getHeight(root->left),getHeight(root->right))+1; } vector> levelOrderBottom(TreeNode *root) { if(!root) return vector >(); vector > res(getHeight(root),vector ()); //初始化二维数组 dfs(root,res.size()-1,res); return res; } void dfs(TreeNode *root,int height,vector > &res) //定义时取应用,避免复制&res { if(!root) return; res[height].push_back(root->val); dfs(root->left,height-1,res); dfs(root->right,height-1,res); } };
4) 思路:用递归实现层序遍历
与正常遍历不同的是,先进行下一层递归,再把当前层的结果保存到res中
//实现1 res定义为私有变量 class Solution {public: vector> levelOrderBottom(TreeNode *root) { // vector > res; if(!root) return res; queue currQueue; currQueue.push(root); levelOrderBottom(currQueue); return res; } void levelOrderBottom(queue currQueue){ if(currQueue.empty()) return; int numLevel = currQueue.size(); //层数 vector row; //读取一层 for(int i=0;i left) currQueue.push(pNode->left); if(pNode->right) currQueue.push(pNode->right); row.push_back(pNode->val); } levelOrderBottom(currQueue); //先递归后存储,递归到最后一行时,才会开始存储。因此会先存最后一行,满足题目倒叙要求 res.push_back(row); }private: vector > res;};
//实现2 res在函数内定义,并传入引用。避免复制引起的操作class Solution {public: vector> levelOrderBottom(TreeNode *root) { vector > res; if(!root) return res; queue currQueue; currQueue.push(root); levelOrderBottom(currQueue,res); return res; } void levelOrderBottom(queue currQueue,vector >& res){ if(currQueue.empty()) return; int numLevel = currQueue.size(); //层数 vector row; //读取一层 for(int i=0;i left) currQueue.push(pNode->left); if(pNode->right) currQueue.push(pNode->right); row.push_back(pNode->val); } levelOrderBottom(currQueue,res); //先递归后存储,递归到最后一行时,才会开始存储。因此会先存最后一行,满足题目倒叙要求 res.push_back(row); }};