Given a binary tree, return the preorder traversal of its nodes‘ values.
Example:
Input:[1,null,2,3]1 2 / 3 Output:[1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?
二叉树的先序遍历。
这个比较简单，直接看递归和非递归的代码实现
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list=new ArrayList<>();
process(root,list);
return list;
}
public void process(TreeNode node,List<Integer> list){
if(node==null){
return;
}
list.add(node.val);
process(node.left,list);
process(node.right,list);
}
}
先序遍历：中左右
非递归的方式用stack来实现(深度优先遍历用stack,广度优先用队列)
对当前节点压入stack，弹出并打印（中）,再对cur压入右左（左右）
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list=new ArrayList<>();
if(root==null){
return list;
}
Stack<TreeNode> stack=new Stack<>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode cur=stack.pop();
list.add(cur.val);
if(cur.right!=null){
stack.push(cur.right);
}
if(cur.left!=null){
stack.push(cur.left);
}
}
return list;
}
}
原文：https://www.cnblogs.com/iwyc/p/15210915.html