preorder, inorder, postorder traversal
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?
Recursive solution
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
helper(root, ret);
return ret;
}
private void helper(TreeNode root, List<Integer> ret) {
if (root == null)
return;
ret.add(root.val);
helper(root.left, ret);
helper(root.right, ret);
}
}
Iterative solution
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
if (root == null)
return ret;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
ret.add(node.val);
if (node.right != null) {
stack.push(node.right);
}
if (node.left != null) {
stack.push(node.left);
}
}
return ret;
}
}
本文详细介绍了如何使用递归和迭代方法实现二叉树的前序遍历,并提供了对应的Java代码实现,旨在提高对二叉树遍历算法的理解和实践能力。
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