94. Binary Tree Inorder Traversal（二叉树中序遍历）

 

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1
    
     2
    /
   3

return [1,3,2].

递归：

class Solution {
    List<Integer> res = new ArrayList<Integer>(); 
    public List<Integer> inorderTraversal(TreeNode root) {
          help(root);
        return res;
    }
    private void help(TreeNode root){
        if(root==null) return ;
        if(root.left!=null)
            help(root.left);
        
        res.add(root.val);
        
        if(root.right!=null)
            help(root.right);
    }
}

非递归：

终极版:

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        Stack<TreeNode> s = new Stack<TreeNode>();
        List<Integer> res = new ArrayList<Integer>();
        while(root!=null||!s.isEmpty()){
            while(root!=null){
                s.push(root);
                root=root.left;
            }
            if(!s.isEmpty()){
                root = s.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}

利用辅助桟

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