binaryTree.preOrderTraversal(); System.out.println(""); binaryTree.inOrderTraversal(); System.out.println(""); binaryTree.postOrderTraversal(); }publicstaticclassBinaryTree {privateNode root;publicvoidpreOrderTraversal() {this.root.preOrderTraversal(); }publicvoidinOrderTraversal() {this.root.in...
二叉树遍历
Map<Integer, List<Integer>> map = new HashMap(); public List<List<Integer>> verticalTraversal(TreeNode root) { List<List<Integer>> res = new ArrayList(); if(root==null) return res; // 借助了队列qt和qi,代替了递归 //树的所有节点 Queue<TreeNode> qt = new LinkedList(); // 节点坐标...
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)defprintPostorder(root):ifroot:printPostorder(root.left)printPostorder(...
If we are given a binary tree and we need to perform a vertical order traversal, that means we will be processing the nodes of the binary tree from left to right. Suppose we have a tree such as the one given below. If we traverse the tree in vertical order and print the nodes then...
a) Make current as right child of the rightmost node in current's left subtree b) Go to this left child, i.e., current = current->left /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; ...
Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? C++ 解法一:
Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 代码语言:javascript 复制 3/\920/\157 return its level order traversal as: ...
Morris Traversal is a method based on the idea of a threaded binary tree and can be used to traverse a tree without using recursion or stack. Morris traversal involves: Step 1:Creating links to inorder successors Step 2:Printing the information using the created links (the tree is altered ...