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1: Pre-sequence traversal (root, left, right)The recursive approach is simple:public static void Pr (TreeNode root) {if (root!=null) {System.out.println (root.val);p R (root.left);p R (root.right); }}Non-recursive method: You can use the stack, the order of the stack is right, left, root.The code is as follows:Public list2: Middle sequence traversal (left, Root, right)Recursive method:public static void M (TreeNode root) {if (root!=null) {m (root.left); System.out.println (Root.val); M (root.rig

=cur->Right ; +Stac.top (). second=2; -Stac.push (Make_pair (cur,0)); the } * ElseStac.pop ();//None of the above two can get in or go through or have no children . $ }Panax Notoginseng returnans; - } the};AC Code(4) More iterative method:1 /**2 * Definition for a binary tree node.3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8 * };9 */Ten classSolution { One Public

Given a binary tree, return the preorder traversal of its nodes 'values. For example:Given Binary Tree{1,#,2,3}, 1 2 / 3 Return[1,2,3]. /** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public ListIdea: Non-recursive forward Traversal

/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: vector Leetcode Binary Tree preorder traversal

Re-build a tree from the central traversal + forward/backward traversal. Iterator must be used before it can be used. Otherwise, MLE is used. 1. preorder + inorder The first version uses the coordinate range: 1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 Tree

[Question] Given a binary tree, return the preorder traversal of its nodes 'values. For example:Given Binary Tree{1,#,2,3}, 1 2 / 3 Return[1,2,3]. Note: recursive solution is trivial, cocould You Do It iteratively? Question] Returns the first-order traversal result without recursion. [Idea] Maintain a stack. [Code] /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right;

; order NBSP; by lft The result is c,d,e,f, and the order is correct. 4, get all the next Level 2 sub-node We have a example, this time with the LVL parameter, because a level is 1, so we query level is not greater than 3. 　　 1Select* fromTreewhereLfT>2 andright One andlvl3Order byLfT Let's look at our new method for adding a node. Below the root node, we add an X node to the right of the G node The job we're going to do is 1, the right parameter of the G node is 13 2, change all the

Given a binary tree, return the preorder traversal of its nodes 'values. For example: Given Binary Tree {1, #, 2, 3 }, 1 2 / 3 Return [1, 2, 3]. Note: Recursive solution is trivial, cocould You Do It iteratively? Idea: Take the top element of the stack, output the element and exit it from the stack. If the right subtree is not empty, add the right subtree to the stack. If the left subtree is not empty, add the left subtree to the stack

Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. Idea: Use the first-order traversal of the provided root node information, find the root node in the middle-order traversal, and then divide the sequence into {left subtree, root, right subtree }, recursively solve left and right subtree. 1 class Solution { 2 public: 3 TreeNode *buildTree( vector

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: TreeNode *buildTree(vector

Given a binary tree, returnPreorderTraversal of its nodes 'values. For example:Given Binary Tree{1,#,2,3}, 1 2 / 3 Return[1,2,3]. Note: recursive solution is trivial, cocould You Do It iteratively? Solution 1: Non-recursive 1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */10 public class Solution {11 public List Solution 2: Recursion 1 /*

Label: des style color Io Java ar strong for div Given a binary tree, returnPreorderTraversal of its nodes 'values. For example:Given Binary Tree{1, #, 2, 3}, 1 2/3 Return[1, 2, 3]. Note:Recursive solution is trivial, cocould You Do It iteratively? Answer /*** Definition for binary tree * Public class treenode {* int val; * treenode left; * treenode right; * treenode (int x) {val = x ;} *} */public class solution {public list Binary Tree pre