Leetcode-545. Boundary of Binary Tree

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。这次比赛略无语，没想到前3题都可以用暴力解。

博客链接：mcf171的博客

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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-mostnode.
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].

这个题目写的有点不优雅，比较通俗易懂的就是先遍历左边界，再遍历子节点，再遍历右边界。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private Set<TreeNode> visited = new HashSet<TreeNode>();
private boolean leftBoundary = true, rightBoundary = true;
public List<Integer> boundaryOfBinaryTree(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
if(root != null){
result.add(root.val);
visited.add(root);
if(root.left != null){
getLeft(root.left, result);
}
if(root.right != null){
getRight(root.right, result, true);
}
}
return result;
}

private void getLeft(TreeNode root, List<Integer> result){
if(root == null)return;
if(leftBoundary){
if(!visited.contains(root)){
result.add(root.val);
visited.add(root);
}
if(root.left == null && root.right == null){leftBoundary = false;return;}

}else{
if(root.left == null && root.right == null && !visited.contains(root)){
result.add(root.val);
visited.add(root);
return;
}
}
getLeft(root.left, result);
getLeft(root.right, result);

}

private void getRight(TreeNode root, List<Integer>result, boolean flag){
if(root == null) return;
if(flag){
if(root.right != null){
getRight(root.left, result, !flag);
getRight(root.right, result, flag);
}else{
getRight(root.left, result, flag);
}
if(!visited.contains(root)){
result.add(root.val);
visited.add(root);
}
}else{
if(root.right == null && root.left == null){
if(!visited.contains(root)){
result.add(root.val);
visited.add(root);
}
}
getRight(root.left, result, flag);
getRight(root.right, result, flag);
}

}
}