[LeetCode] 545. Boundary of Binary Tree 二叉树的边界

2021年09月15日 阅读数:1
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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.html

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.java

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.node

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

Example 1app

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 2post

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].

给了一个二叉树,以逆时针顺序来输出树的边界,按顺序分别为左边界,叶结点和右边界。this

对于左边界,叶节点和右边界分别用递归或者迭代二叉树的节点求3次。url

解法1: 递归htm

解法2: 迭代blog

Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
      
    public List<Integer> boundaryOfBinaryTree(TreeNode root) {
        List<Integer>list=new ArrayList<>();
        if(root==null){
            return list;
        }
        list.add(root.val);
        
        GetLeftPath(root.left,list);//add left boundary node and leaves node
        GetRightPath(root.right, list);// add right boundary node and leaves node
      
        return list;
    }
    
    public void GetLeftPath(TreeNode left,List<Integer>list){
        if(left!=null){
            list.add(left.val);// add the left boundary node
            if(left.left!=null){
                GetLeftPath(left.left, list);
                DFS(left.right,list);
            }
            else{// according to the rule, if the node has no left subtree,then the left path goes to right
                GetLeftPath(left.right, list);
            }
        }
    }
    
    public void GetRightPath(TreeNode right,List<Integer>list){
        if(right!=null){
            if(right.right!=null){
                DFS(right.left,list);
                GetRightPath(right.right, list);
            }
            else{
                //according to the rule,if the node has no right subtree,then the right path goes to left
                GetRightPath(right.left, list);
            }
            list.add(right.val);
        }
    }
    
 
    public void DFS(TreeNode node,List<Integer>list){
        if(node!=null){
            if(node.left==null&&node.right==null){
                list.add(node.val);
            }
            else{
                DFS(node.left, list);
                DFS(node.right,list);
            }
        }
    }
}  

Python:

# Time:  O(n)
# Space: O(h)

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution(object):
    def boundaryOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        def leftBoundary(root, nodes):
            if not root or (not root.left and not root.right):
                return
            nodes.append(root.val)
            if not root.left:
                leftBoundary(root.right, nodes)
            else:
                leftBoundary(root.left, nodes)

        def rightBoundary(root, nodes):
            if not root or (not root.left and not root.right):
                return
            if not root.right:
                rightBoundary(root.left, nodes)
            else:
                rightBoundary(root.right, nodes)
            nodes.append(root.val)

        def leaves(root, nodes):
            if not root:
                return
            if not root.left and not root.right:
                nodes.append(root.val)
                return
            leaves(root.left, nodes)
            leaves(root.right, nodes)

        if not root:
            return []

        nodes = [root.val]
        leftBoundary(root.left, nodes)
        leaves(root.left, nodes)
        leaves(root.right, nodes)
        rightBoundary(root.right, nodes)
        return nodes

C++:

class Solution {
public:
    vector<int> boundaryOfBinaryTree(TreeNode* root) {
        if (!root) return {};
        vector<int> res;
        if (root->left || root->right) res.push_back(root->val);
        leftBoundary(root->left, res);
        leaves(root, res);
        rightBoundary(root->right, res);
        return res;
    }
    void leftBoundary(TreeNode* node, vector<int>& res) {
        if (!node || (!node->left && !node->right)) return;
        res.push_back(node->val);
        if (!node->left) leftBoundary(node->right, res);
        else leftBoundary(node->left, res);
    }
    void rightBoundary(TreeNode* node, vector<int>& res) {
        if (!node || (!node->left && !node->right)) return;
        if (!node->right) rightBoundary(node->left, res);
        else rightBoundary(node->right, res);
        res.push_back(node->val);
    }
    void leaves(TreeNode* node, vector<int>& res) {
        if (!node) return;
        if (!node->left && !node->right) {
            res.push_back(node->val);
        }
        leaves(node->left, res);
        leaves(node->right, res);
    }
};

C++:

class Solution {
public:
    vector<int> boundaryOfBinaryTree(TreeNode* root) {
        if (!root) return {};
        vector<int> res{root->val};
        helper(root->left, true, false, res);
        helper(root->right, false, true, res);
        return res;
    }
    void helper(TreeNode* node, bool leftbd, bool rightbd, vector<int>& res) {
        if (!node) return;
        if (!node->left && !node->right) {
            res.push_back(node->val);
            return;
        }
        if (leftbd) res.push_back(node->val);
        helper(node->left, leftbd && node->left, rightbd && !node->right, res);
        helper(node->right, leftbd && !node->left, rightbd && node->right, res);
        if (rightbd) res.push_back(node->val);
    }
};

C++:

class Solution {
public:
    vector<int> boundaryOfBinaryTree(TreeNode* root) {
        if (!root) return {};
        vector<int> res, right;
        TreeNode *l = root->left, *r = root->right, *p = root;
        if (root->left || root->right) res.push_back(root->val);
        while (l && (l->left || l->right)) {
            res.push_back(l->val);
            if (l->left) l = l->left;
            else l = l->right;
        }
        stack<TreeNode*> st;
        while (p || !st.empty()) {
            if (p) {
                st.push(p);
                if (!p->left && !p->right) res.push_back(p->val);
                p = p->left;
            } else {
                p = st.top(); st.pop();
                p = p->right;
            }
        }
        while (r && (r->left || r->right)) {
            right.push_back(r->val);
            if (r->right) r = r->right;
            else r = r->left;
        }
        res.insert(res.end(), right.rbegin(), right.rend());
        return res;
    }
};

  

 

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