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, could you do it iteratively?
中文:二叉树的前序遍历(根-左-右)。
能用非递归实现吗?
递归:
public class BinaryTreePreorderTraversal { public List<Integer> preorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); if(root == null) return list; list.add(root.val); list.addAll(preorderTraversal(root.left)); list.addAll(preorderTraversal(root.right)); return list; } // Definition for binary tree public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } } }非递归:先把右节点的值压入栈中,再压入左的。弹出左的,弹出右的……。
public List<Integer> preorderTraversal(TreeNode root){ List<Integer> list = new ArrayList<Integer>(); if(root == null) return list; Stack<TreeNode> stack = new Stack<TreeNode>(); stack.push(root); while(!stack.isEmpty()){ TreeNode node = stack.pop(); list.add(node.val); if(node.right != null) stack.push(node.right); if(node.left != null) stack.push(node.left); } return list; }