LeetCode 545. Boundary of Binary Tree 二叉树边界

LeetCode 545. Boundary of Binary Tree 二叉树边界
Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes.

Left boundary is defined as the path from root to the left-most node. Right boundary is defined as the path from root to the right-most node. If the root doesn't have left subtree or right subtree, then the root itself is left boundary or right boundary. Note this definition only applies to the input binary tree, and not applies to any subtrees.

The left-most node is defined as a leaf node you could reach when you always firstly travel to the left subtree if exists. If not, travel to the right subtree. Repeat until you reach a leaf node.

The right-most node is also defined by the same way with left and right exchanged.

Example 1
Input:

  1
   \
    2
   / \
  3   4

Ouput:
[1, 3, 4, 2]

Explanation:
The root doesn't have left subtree, so the root itself is left boundary.
The leaves are node 3 and 4.
The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed right boundary.
So order them in anti-clockwise without duplicates and we have [1,3,4,2].

Example 2

Input:

    ____1_____
   /          \
  2            3
 / \          / 
4   5        6   
   / \      / \
  7   8    9  10  

Ouput:
[1,2,4,7,8,9,10,6,3]

Explanation:
The left boundary are node 1,2,4. (4 is the left-most node according to definition)
The leaves are node 4,7,8,9,10.
The right boundary are node 1,3,6,10. (10 is the right-most node).
So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].

题意: 高频题，必须熟练掌握。逆时针打印二叉树边界。

解题思路：
根据观察，我们发现

1. 当node为leftBound左边界时，node.left也是左边界
2. 当node为leftBound左边界时，node.left为空，则node.right也可以leftBound左边界。
3. Bottom的所有都要加入其中。
4. rightBound也是如此。

我们可以循环调用dfs，初始化leftBound和rightBound两个boolean参数，一层层判断。先加入左边，加入bottom，然后得到两个子树加入，最后加入右边界。

代码如下：

/**
    * node.left is left bound if node is left bound;
    * node.right could also be left bound if node is left bound && node has no left child;
    * Same applys for right bound;
    * if node is left bound, add it before 2 child - pre order;
    * if node is right bound, add it after 2 child - post order;
    * A leaf node that is neither left or right bound belongs to the bottom line;
    */
    public List<Integer> boundaryOfBinaryTree(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        res.add(root.val);
        getBounds(root.left, res, true, false);
        getBounds(root.right, res, false, true);
        return res;
    }
    public void getBounds(TreeNode node, List<Integer> res, boolean leftBound, boolean rightBound) {
        if (node == null) return;
        if (leftBound) {
            res.add(node.val);
        }
        //add bottom
        if(!leftBound && !rightBound && node.left == null && node.right == null) {
            res.add(node.val);
        }
        getBounds(node.left, res, leftBound, rightBound && node.right == null);
        getBounds(node.right, res, leftBound && node.left == null, rightBound);
        if (rightBound) {
            res.add(node.val);
        }
        
        
    }