leetcode 144. Binary Tree Preorder 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?

和in-order相比输出位置变一下即可

public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(cur!=null || !stack.empty()){
            while(cur!=null){
                stack.push(cur);
                ans.add(cur.val);
                cur = cur.left;
            }
            cur = stack.pop();
            cur = cur.right;
        }
        return ans;
    }
}

这个过程可以只把右节点放在stack当中

public List<Integer> preorderTraversal(TreeNode node) {
	List<Integer> list = new LinkedList<Integer>();
	Stack<TreeNode> rights = new Stack<TreeNode>();
	while(node != null) {
		list.add(node.val);
		if (node.right != null) {
			rights.push(node.right);
		}
		node = node.left;
		if (node == null && !rights.isEmpty()) {
			node = rights.pop();
		}
	}
    return list;
}

Morris方法在输出的位置要调整一个地方

public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        TreeNode cur=root, pre=null;
        while(cur!=null){
            if(cur.left==null){
                ans.add(cur.val);
                cur = cur.right;
            }else{
                pre = cur.left;
                while(pre.right!=null && pre.right!=cur){
                    pre = pre.right;
                }
                if(pre.right==null){
                    pre.right = cur;
                    ans.add(cur.val);
                    cur = cur.left;
                }else{
                    pre.right = null;
                    cur = cur.right;
                }
                
            }
        }
        return ans;
    }
}