vector<int> postorderTraversal(TreeNode *root) { vector<int>result;if(root==NULL)returnresult; stack<TreeNode*>pTree; pTree.push(root);while(!pTree.empty()) { TreeNode*curNode=pTree.top(); result.push_back(curNode->val); pTree.pop();if(curNode->left) pTree.push(curNode->left);if(curNode->right) pTree.push(curNode->right);...
Given a binary tree, return thepostordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[3,2,1]. Note:Recursive solution is trivial, could you do it iteratively? ++++++++++++++++++++++++++++++++++++++++++ 【二叉树遍历模版】后...
Given a binary tree, return the postorder traversal of its nodes’ values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note:Recursive solution is trivial, could you do it iteratively? 本题难度Hard。有3种算法分别是: 递归...
publicList<Integer>postorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();if(root==null){returnlist;}Stack<TreeNode>stack=newStack<>();stack.push(root);stack.push(root);while(!stack.isEmpty()){TreeNodecur=stack.pop();if(cur==null){continue;}if(!stack.isEmpty()&&cur==stac...
145. Binary Tree Postorder Traversal 二叉树的后序遍历 给定一个二叉树,返回它的后序遍历。 示例: 输入:[1,null,2,3] 1 \ 2 / 3输出:[3,2,1] 1. 2. 3. 4. 5. 6. 7. 8. 进阶:递归算法很简单,你可以通过迭代算法完成吗? DFS 今天这道题比较简单,直接DFS就可以了。
LeetCode 145 题, Binary Tree Postorder Traversal 解题思路。 1、先读题,题目要求后序遍历,用非递归解决。 递归解决很容易,代码如下: # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x ...
postorder traversal sequencebinary search treesbinary tree structureaverage depth/ C4240 Programming and algorithm theory C1160 Combinatorial mathematicsBinary search trees built from the postorder traversal sequence of other binary search trees are characterized in terms of their binary tree structure. A ...
binary-tree-postorder-traversal title: binary-tree-postorder-traversal 描述 Given a binary tree, return thepostordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[3,2,1]. Note:Recursive solution is trivial, could you do it iteratively?
Solution{public:vector<int>postorderTraversal(TreeNode*root){vector<int>result;stack<TreeNode*>st;TreeNode*p=root,*q=NULL;do{while(p)st.push(p),p=p->left;q=NULL;while(!st.empty()){p=st.top();st.pop();if(p->right==q)//右子树为空或者已经访问{result.push_back(p->val);q=p...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...