545.Boundary-of-Binary-Tree

545. Boundary of Binary Tree

题目地址

https://leetcode.com/problems/boundary-of-binary-tree/

题目描述

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.  (The values of the nodes may still be duplicates.)

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].

代码

Approach 1: Simple Solution

Complexity Analysis

• Time complexity : O_(_n) One complete traversal for leaves and two traversals upto depth of binary tree for left and right boundary.

• Space complexity : O(_n) res_res _and stack_stack* is used.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
  public List<Integer> boundaryOfBinaryTree(TreeNode root) {
    ArrayList<Integer> res = new ArrayList<>();
    if (root == null)  return res;

    // add root
    if (!isLeaf(root)) {
      res.add(root.val);
    }

    // add left boundary
    TreeNode t = root.left;
    while (t != null) {
      if (!isLeaf(t)) {
        res.add(t.val);
      }
      if (t.left != null) {
        t = t.left;
      } else {
        t = t.right;
      }
    }

    // add leafs
    addLeaves(res, root);

    // add right boundary with stack
    Stack<Integer> s = new Stack<>();
    t = root.right;
    while (t != null) {
      if (!isLeaf(t)) {
        s.push(t.val);
      }
      if (t.right != null) {
        t = t.right;
      } else {
        t = t.left;
      }
    }
    while (!s.empty()) {
      res.add(s.pop());
    }

    return res;
  }

  public void addLeaves(List<Integer> res, TreeNode root) {
    if (isLeaf(root)) {
        res.add(root.val);
    } else {
      if (root.left != null) {
        addLeaves(res, root.left);
      }
      if (root.right != null) {
        addLeaves(res, root.right);
      }
    }
  }

  public boolean isLeaf(TreeNode t) {
      return t.left == null && t.right == null;
  }
}

Approach #2: Using PreOrder Traversal

public class Solution {
    enum Flag {ROOT, LEFT, RIGHT, MIDDLE};
    public List<Integer> boundaryOfBinaryTree(TreeNode root) {
        if (root == null)     return new ArrayList<>();

        List<Integer> leftBoundary = new ArrayList<>();
        List<Integer> leaves = new ArrayList<>();
        List<Integer> rightBoundary = new ArrayList<>();
        preOrder(root, leftBoundary, leaves, rightBoundary, Flag.ROOT);

        leftBoundary.addAll(leaves);
        leftBoundary.addAll(rightBoundary);
        return leftBoundary;
    }

    // root -> left -> right
    private void preOrder(TreeNode root, List<Integer> leftBoundary, List<Integer> leaves, List<Integer> rightBoundary, Flag flag) {
        if (root == null)         return;

        // Root
        if (flag == Flag.ROOT || flag == Flag.LEFT) {
            leftBoundary.add(root.val);
        } else if (flag == Flag.RIGHT) {
            rightBoundary.add(0, root.val);    // Reverse add
        } else if (root.left == null && root.right == null) {
            leaves.add(root.val);
        }

        // Left
        if (root.left != null) {
            preOrder(root.left, leftBoundary, leaves, rightBoundary, childFlag(root, flag, true));
        }

        //Right
        if (root.right != null) {
            preOrder(root.right, leftBoundary, leaves, rightBoundary, childFlag(root, flag, false));
        }
    }

    private Flag childFlag(TreeNode parent, Flag flag, boolean isLeftChild) {
        if (flag == Flag.ROOT) {
            return isLeftChild ? Flag.LEFT : Flag.RIGHT;
        }

        //Parent on left boundary, if exist left child, the right child should be a middle node
        if (flag == Flag.LEFT && !isLeftChild && parent.left != null) {
            return Flag.MIDDLE;
        }

        //Parent on right boundary, if exist right node, the left child should be a middle node.
        if (flag == Flag.RIGHT && isLeftChild && parent.right != null) {
            return Flag.MIDDLE;
        }

        return flag;
    }
}