二叉树的迭代遍历统一写法

94.二叉树的中序遍历

具体实现：

将访问的节点直接加入到栈中，但如果是出去过的节点则后面再放入一个空节点，

这样只有空节点弹出的时候，才将下一个节点放进结果集。

代码：

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new LinkedList<>();
        TreeNode cur = root;
        if(cur != null) stack.push(cur);
        while (!stack.isEmpty()) {
            cur = stack.peek();
            if (cur != null) {
                cur = stack.pop();
                if (cur.right != null) stack.push(cur.right);
                stack.push(cur);
                stack.push(null);
                if (cur.left != null) stack.push(cur.left);
            } else {
                stack.pop();
                cur = stack.pop();
                result.add(cur.val);
            }
        }
        return result;
    }
}

144.二叉树的前序遍历

代码：

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new LinkedList<>();
        TreeNode cur = root;
        if(cur != null) stack.push(cur);
        while (!stack.isEmpty()) {
            cur = stack.peek();
            if (cur != null) {
                cur = stack.pop();
                if (cur.right != null) stack.push(cur.right);
                if (cur.left != null) stack.push(cur.left);
                stack.push(cur);
                stack.push(null);
            } else {
                stack.pop();
                cur = stack.pop();
                result.add(cur.val);
            }
        }
        return result;
    }
}

145.二叉树的后序遍历

代码：

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new LinkedList<>();
        TreeNode cur = root;
        if(cur != null) stack.push(cur);
        while (!stack.isEmpty()) {
            cur = stack.peek();
            if (cur != null) {
                cur = stack.pop();
                stack.push(cur);
                stack.push(null);
                if (cur.right != null) stack.push(cur.right);
                if (cur.left != null) stack.push(cur.left);
                
            } else {
                stack.pop();
                cur = stack.pop();
                result.add(cur.val);
            }
        }
        return result;
    }
}