Given a binary tree, return the inorder traversal of its nodes' values.
Example:
Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2]
Follow up: Recursive solution is trivial, could you do it iteratively?
public List<Integer> inorderTraversal(TreeNode root) {// 树的中序遍历 递归 my
List<Integer> res = new ArrayList<Integer>();
inorderT(root,res);
return res;
}
private void inorderT(TreeNode root,List list){
if(null!=root){
inorderT(root.left,list);
list.add(root.val);
inorderT(root.right,list);
}
}
public List<Integer> inorderTraversal(TreeNode root) {//树的中序遍历 迭代 myTip
List<Integer> res = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode cur = root;
while(null!= cur || !stack.isEmpty()){
while(null!=cur){
stack.push(cur);
cur=cur.left;
}
cur= stack.pop();
res.add(cur.val);
cur= cur.right;
}
return res;
}