LF.44.Pre-order Traversal Of Binary Tree

Implement an iterative, pre-order traversal of a given binary tree, return the list of keys of each node in the tree as it is pre-order traversed.

Examples

5

/

3 8

/

1 4 11

Pre-order traversal is [5, 3, 1, 4, 8, 11]

Corner Cases

What if the given binary tree is null? Return an empty list in this case.

tree 结构的 空间复杂度，都需要考虑树的结构，o（h）是正确答案。

interative: 时间复杂度： o(n) 每个点都走， 空间复杂度 o(h) 但是是在 heap 上

public class Solution {
  public List<Integer> preOrder(TreeNode root) {
    // Write your solution here
    List<Integer> res = new ArrayList<>();
    if (root == null) {
        return res ;
    }
    Deque<TreeNode> stack = new LinkedList<>() ;
    stack.offerFirst(root) ;
    while(!stack.isEmpty()){
        TreeNode curr = stack.pollFirst() ;
        res.add(curr.key);
        if (curr.right != null) {
            stack.offerFirst(curr.right);
        }
        if (curr.left != null) {
            stack.offerFirst(curr.left);
        }
    }
    return res ;
  }
}

 recursive : time: o(n) for every node, space: o(h) on the stack: 

private List<Integer> preOrder_recur(TreeNode root){
        List<Integer> res = new ArrayList<>() ;
        if (root == null) {
            return res ;
        }
        helper(root, res);
        return res ;
    }
    private void helper(TreeNode root , List<Integer> res ){
            //base case
        if (root == null) {
            return ;
        }
        res.add(root.key) ;
        helper(root.left, res) ;
        helper(root.right, res);
    }