144. 二叉树的前序遍历 94. 二叉树的中序遍历 145. 二叉树的后序遍历

给定一个二叉树，返回它的 前序 遍历。

 示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [1,2,3]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

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;
    }
}

class Solution {
    List<Integer> list = new ArrayList();
    public List<Integer> preorderTraversal(TreeNode root) {
        if(root == null)
            return list;
        list.add(root.val);
        preorderTraversal(root.left);
        preorderTraversal(root.right);
        return list;
    }
}

---

给定一个二叉树，返回它的中序 遍历。

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList();
        if(root == null)
            return list;
        Stack<TreeNode> stack = new Stack();
        TreeNode p = root;
        while(!stack.isEmpty() || p != null){
            while(p != null){//注意这里
                stack.push(p);
                p = p.left;
            }
            p = stack.pop();
            list.add(p.val);
            p = p.right;//注意这里
        }
        return list;
    }
}

---

给定一个二叉树，返回它的 后序 遍历。

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList();
        if(root == null)
            return list;
        Stack<TreeNode> stack1 = new Stack();
        Stack<TreeNode> stack2 = new Stack();
        stack1.push(root);
        while(!stack1.isEmpty()){
            TreeNode cur = stack1.pop();
            stack2.push(cur);
            if(cur.left != null){
                stack1.push(cur.left);
            }
            if(cur.right != null)
                stack1.push(cur.right);
        }
        while(!stack2.isEmpty()){
            TreeNode cur = stack2.pop();
            list.add(cur.val);
        }
        return list;
    }
}

或者使用： Collections.reverse(list);

class Solution {
    public List<Integer> postorderTraversal(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.left != null){
                stack.push(cur.left);
            }
            if(cur.right != null)
                stack.push(cur.right);
        }
        Collections.reverse(list);
        return list;
    }
}