*/classSolution{public List<Integer>inorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){// left firstif(current.left==null){result.add(current.val);current=current.right;}// if there is left, get the rightmost ...
中序Inorder: 先访问左子树,然后访问根节点,最后访问右子树. 后序Postorder:先访问左子树,然后访问右子树,最后访问根节点. classNode:def__init__(self,key):self.left=Noneself.right=Noneself.val=keydefprintInorder(root):ifroot:printInorder(root.left)print(root.val)printInorder(root.right)defprintP...
int[]postorder){inIndex=inorder.length-1;postIndex=postorder.length-1;returnpostBuildTree(inorder,postorder,null);}privateTreeNodepostBuildTree(int[]inorder,int[]postorder,TreeNoderoot){if(postIndex<0){returnnull;}TreeNoden=newTreeNode(postorder[postIndex--]);// find rightest node// build ...
3.Post Order Traverse 1publicList<Integer>postorderTraversal(TreeNode root) {2LinkedList<Integer> result =newLinkedList<>();3Deque<TreeNode> stack =newArrayDeque<>();4TreeNode p =root;5while(!stack.isEmpty() || p !=null) {6if(p !=null) {7stack.push(p);8result.addFirst(p.val);...
postorderTraversal(result, root.right); result.add(root.val); } public void inorderTraversal(List<Integer> result, TreeNode root){ if(root==null) return; inorderTraversal(result, root.left); result.add(root.val); inorderTraversal(result, root.right); ...
preorderinorderpostorder""" :type preorder: List[int] :type inorder: List[int] :type postorder: List[int] :rtype: bool """ifnotpreorderornotinorderornotpostorder:returnFalseifset(preorder)!=set(inorder)orset(preorder)!=set(postorder):returnFalseiflen(preorder)!=len(inorder)orlen(in...
<< endl; tree->Inorder(tree->Root()); cout << endl; cout << "Post order traversal" << endl; tree->Postorder(tree->Root()); cout << endl; delete tree; return 0; } Edit & run on cpp.shTopic archived. No new replies allowed.Home...
preorder: root-left-rightinorder: left-root-rightpostorder: left-right-root
这段代码中的 print_post_order 函数从 Inorder 和 Preorder 遍历打印 Postorder 遍历。如果 Inorder 和 Preorder 为空,则返回。否则,通过 Preorder 中的第一个节点选取根节点,并在 Inorder 中查找该根节点的位置和左右子树。然后,我们递归地处理在该节点左侧和右侧的 Inorder,以及在 Preorder 中与这些 Inord...
Is there a way to print out numbers for pre-order, post-order and in-order traversals of trees using just recursion and a number. The trees are binary and n is the numbe