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1361. Validate Binary Tree Nodes

Description

You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.

If node i has no left child then leftChild[i] will equal -1, similarly for the right child.

Note that the nodes have no values and that we only use the node numbers in this problem.

 

Example 1:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
Output: true

Example 2:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
Output: false

Example 3:

Input: n = 2, leftChild = [1,0], rightChild = [-1,-1]
Output: false

 

Constraints:

  • n == leftChild.length == rightChild.length
  • 1 <= n <= 104
  • -1 <= leftChild[i], rightChild[i] <= n - 1

Solutions

Solution 1: Union-Find

We can traverse each node $i$ and its corresponding left and right children $l$, $r$, using an array $vis$ to record whether the node has a parent:

  • If the child node already has a parent, it means there are multiple fathers, which does not meet the condition, so we return false directly.
  • If the child node and the parent node are already in the same connected component, it means a cycle will be formed, which does not meet the condition, so we return false directly.
  • Otherwise, we perform a union operation, set the corresponding position of the $vis$ array to true, and decrease the number of connected components by $1$.

After the traversal, we check whether the number of connected components in the union-find set is $1$. If it is, we return true, otherwise, we return false.

The time complexity is $O(n \times \alpha(n))$, and the space complexity is $O(n)$. Where $n$ is the number of nodes, and $\alpha(n)$ is the inverse Ackermann function, which is less than $5$.

  • class Solution {
        private int[] p;
    
        public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
            p = new int[n];
            for (int i = 0; i < n; ++i) {
                p[i] = i;
            }
            boolean[] vis = new boolean[n];
            for (int i = 0, m = n; i < m; ++i) {
                for (int j : new int[] {leftChild[i], rightChild[i]}) {
                    if (j != -1) {
                        if (vis[j] || find(i) == find(j)) {
                            return false;
                        }
                        p[find(i)] = find(j);
                        vis[j] = true;
                        --n;
                    }
                }
            }
            return n == 1;
        }
    
        private int find(int x) {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        }
    }
    
  • class Solution {
    public:
        bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
            int p[n];
            iota(p, p + n, 0);
            bool vis[n];
            memset(vis, 0, sizeof(vis));
            function<int(int)> find = [&](int x) {
                return p[x] == x ? x : p[x] = find(p[x]);
            };
            for (int i = 0, m = n; i < m; ++i) {
                for (int j : {leftChild[i], rightChild[i]}) {
                    if (j != -1) {
                        if (vis[j] || find(i) == find(j)) {
                            return false;
                        }
                        p[find(i)] = find(j);
                        vis[j] = true;
                        --n;
                    }
                }
            }
            return n == 1;
        }
    };
    
  • class Solution:
        def validateBinaryTreeNodes(
            self, n: int, leftChild: List[int], rightChild: List[int]
        ) -> bool:
            def find(x: int) -> int:
                if p[x] != x:
                    p[x] = find(p[x])
                return p[x]
    
            p = list(range(n))
            vis = [False] * n
            for i, (a, b) in enumerate(zip(leftChild, rightChild)):
                for j in (a, b):
                    if j != -1:
                        if vis[j] or find(i) == find(j):
                            return False
                        p[find(i)] = find(j)
                        vis[j] = True
                        n -= 1
            return n == 1
    
    
  • func validateBinaryTreeNodes(n int, leftChild []int, rightChild []int) bool {
    	p := make([]int, n)
    	for i := range p {
    		p[i] = i
    	}
    	var find func(int) int
    	find = func(x int) int {
    		if p[x] != x {
    			p[x] = find(p[x])
    		}
    		return p[x]
    	}
    	vis := make([]bool, n)
    	for i, a := range leftChild {
    		for _, j := range []int{a, rightChild[i]} {
    			if j != -1 {
    				if vis[j] || find(i) == find(j) {
    					return false
    				}
    				p[find(i)] = find(j)
    				vis[j] = true
    				n--
    			}
    		}
    	}
    	return n == 1
    }
    

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