代码随想录刷题笔记-Day12

1. 二叉树的递归遍历

144. 二叉树的前序遍历https://leetcode.cn/problems/binary-tree-preorder-traversal/94. 二叉树的中序遍历https://leetcode.cn/problems/binary-tree-inorder-traversal/145. 二叉树的后续遍历https://leetcode.cn/problems/binary-tree-postorder-traversal/二叉树的深度优先遍历，是一个递归式的算法，对于递归，需要注意的就是每次递归的返回值，以及每次递归需要的参数，以及中止条件和进入递归的判断逻辑。

代码

• 前序

class Solution {
	public List preorderTraversal(TreeNode root) {
		List list = new ArrayList<>();
		pre(root, list);
		return list;
	}

	private void pre(TreeNode root, List result) {
		if (root == null)
			return;
		result.add(root.val);
		pre(root.left, result);
		pre(root.right, result);
	}
}

• 中序

class Solution {
    public List inorderTraversal(TreeNode root) {
        List list = new ArrayList<>();
        inorder(root, list);
        return list;
    }

    private void inorder(TreeNode root, List result) {
        if (root == null)
            return;
        inorder(root.left, result);
        result.add(root.val);
        inorder(root.right, result);
    }
}

•  后序

class Solution {
    public List postorderTraversal(TreeNode root) {
        List list = new ArrayList<>();
        postorder(root, list);
        return list;
    }

    private void postorder(TreeNode root, List result) {
        if (root == null)
            return;
        postorder(root.left, result);
        postorder(root.right, result);
        result.add(root.val);
    }
}

 2. 二叉树的迭代遍历

使用栈进行迭代遍历

• 前序

class Solution {
	public List preorderTraversal(TreeNode root) {
		List result = new ArrayList<>();
		if (root == null)
			return result;
		Stack stack = new Stack<>();

		stack.push(root);
		while (!stack.isEmpty()) {
			TreeNode node = stack.pop();
			result.add(node.val);
			if (node.right != null) {
				stack.push(node.right);
			}
			if (node.left != null) {
				stack.push(node.left);
			}
		}

		return result;
	}
}

• 中序

 中序遍历需要注意：

        中序遍历是先一路找到最左的，然后输出，每次输出就是处理一个node，输出后就得找到右节点的最左的。 

class Solution {
    public List inorderTraversal(TreeNode root) {
        List result = new ArrayList<>();
        if (root == null)
            return result;
        Stack stack = new Stack<>();

        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            if (cur != null) {
                stack.push(cur);
                cur = cur.left;
            } else {
                cur = stack.pop();
                result.add(cur.val);
                cur = cur.right;
            }
        }

        return result;
    }
}

• 后序

class Solution {
    public List postorderTraversal(TreeNode root) {
        List result = new ArrayList<>();
        if (root == null)
            return result;
        Stack stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            result.add(node.val);
            if (node.left != null) {
                stack.push(node.left);
            }
            if (node.right != null) {
                stack.push(node.right);
            }
        }
        Collections.reverse(result);
        return result;
    }
}