# Boundary of Binary Tree

Leave a commentMay 26, 2017 by oneOokay

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 4Ouput:[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 10Ouput:[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].

这题神奇了…简直是对二分法 recusrion用到了出神入化了….

从上到下的到最左儿子, leaves, 从下到上的最右儿子到root.

不能存在重复的值,那么首先要分割好问题.分三个部分:

- 加入从上到下到最左儿子: 除了它的最后一个元素,因为他是一个leaf. 因为对leaf比较较好判断.
- 加入所有的leaves
- 加入从下到上的从最右儿子到root:除了它的第一个元素, 因为他是一个leaf

private List<Integer> ans; public List boundaryOfBinaryTree(TreeNode root) { ans = new ArrayList<Integer>(); if (root == null) return ans; //先加入root ans.add(root.val); //从root的左儿子开始进行left boundary leftboundary(root.left); //从root的左儿子开始找叶子 leaves(root.left); //从root的右儿子开始找叶子 leaves(root.right); //从root的右儿子开始进行right boundary rightboundary(root.right); return ans; } private void leftboundary(TreeNode node){ //如果当前node为null或者node为叶子的话,不加入肯德基豪华套餐 if (node == null || (node.left == null && node.right ==null)) return; //加入.所以这里是按从上到下的顺序按recursion的路径加 ans.add(node.val); //注意这里就是转个弯继续找最左儿子 if (node.left == null) leftboundary(node.right); else leftboundary(node.left); } private void rightboundary(TreeNode node){ if (node == null || (node.left == null && node.right == null)) return; if (node.right == null) rightboundary(node.left); else rightboundary(node.right); //在recursion最终加入ans,表示是reverse order ans.add(node.val); } //如何从左到右求所有的leaves private void leaves(TreeNode node){ if (node == null) return; if (node.left == null && node.right == null) { ans.add(node.val); return; } //先左后右 leaves(node.left); leaves(node.right); }