Binary Tree Preorder Traversal

Binary Tree Preorder Traversal

Problem:

Given a binary tree, return the preorder traversal of its nodes ' values.

Recursive solution is trivial, could do it iteratively?

Ideas:

Stack method pops up the top of the stack when a node is added to do the node until the stack empty

My Code:

 Public classSolution { PublicList<integer>preordertraversal (TreeNode root) {List<Integer> list =NewArraylist<integer>(); if(Root = =NULL)returnlist; Stack<TreeNode> stack =NewStack<treenode>();        Stack.push (root);  Do{TreeNode node=Stack.pop ();            List.add (Node.val); if(Node.right! =NULL) Stack.push (node.right); if(Node.left! =NULL) Stack.push (node.left); } while(!stack.isempty ()); returnlist; }}

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Others code:

 Public classSolution { PublicList<integer>preordertraversal (TreeNode root) {Stack<TreeNode> stack =NewStack<treenode>(); List<Integer> preorder =NewArraylist<integer>(); if(Root = =NULL) {            returnpreorder;        } stack.push (root);  while(!Stack.empty ()) {TreeNode node=Stack.pop ();            Preorder.add (Node.val); if(Node.right! =NULL) {Stack.push (node.right); }            if(Node.left! =NULL) {Stack.push (node.left); }        }                returnpreorder; }}

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The Learning Place:

• A binary tree is a tree with two forks finding the tree is satisfying the left subtree < node < right subtree two fork balance tree Saozi The height difference between the right sub-tree is 1

Binary Tree Preorder Traversal