* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> inorderTraversal(TreeNode*root) { vector<int>ret;if( !root )returnret; stack<TreeNode*>sta; sta.push(root);while( !sta.empty() ) { TreeNode*tmp =sta.top(); sta.pop();if(...
inorderTraversal(root.right); } return list; } } 非递归实现: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> inorderTravers...
vector<int> inorderTraversal(TreeNode* root) { vector<int> res; stack<TreeNode*> treeroot; while(root!=NULL||!treeroot.empty()) { while(root!=NULL) { treeroot.push(root); root=root->left; } root = (); treeroot.pop(); res.push_back(root->val); root = root->right; } retu...
94. Binary Tree Inorder Traversal Given a binary tree, return the inorder traversal of its nodes’ values. For example: Given binary tree [1,null,2,3], AI检测代码解析 \ 1. return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively?
Binary Tree Level Order Traversal 二叉树层序遍历 Example Givenbinary tree[3,9,20,null,null,15,7],3/\920/\157returnits level order traversalas:[[3],[9,20],[15,7]] BFS方法 var levelOrder=function(root){if(!root)return[]conststack=[root]constres=[]while(stack.length){constlen=stack...
名字叫做, morrois traversal, 自己写了下: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>inorderTraversal(TreeNoderoot){List...
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...
The preorder traversal sequence of a binary tree is abdgcefh, the inorder traversal sequence of a binary tree is dgbaechf, the postorder traversal sequence of a binary tree is ___.A.bdgcefhaB.gdbecfhaC.bdgaechfD.gdbehfca的答案是什么.用刷刷题APP,拍照搜索
解法三 Morris Traversal 解法一和解法二本质上是一致的,都需要 O(h)的空间来保存上一层的信息。而我们注意到中序遍历,就是遍历完左子树,然后遍历根节点。如果我们把当前根节点存起来,然后遍历左子树,左子树遍历完以后回到当前根节点就可以了,怎么做到呢?
题目Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], return [1,3,2]. Note: Recursive solution is trivial, could you do it iter…