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993. Cousins in Binary Tree

Description

Given the root of a binary tree with unique values and the values of two different nodes of the tree x and y, return true if the nodes corresponding to the values x and y in the tree are cousins, or false otherwise.

Two nodes of a binary tree are cousins if they have the same depth with different parents.

Note that in a binary tree, the root node is at the depth 0, and children of each depth k node are at the depth k + 1.

 

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

 

Constraints:

  • The number of nodes in the tree is in the range [2, 100].
  • 1 <= Node.val <= 100
  • Each node has a unique value.
  • x != y
  • x and y are exist in the tree.

Solutions

Solution 1

BFS or DFS.

Solution 2

We design a function $dfs(root, parent, depth)$, which means starting from the root node $root$, whose parent node is $parent$, and the depth is $depth$, a depth-first search is performed.

In the function, we first determine whether the current node is empty, and if it is empty, return it directly. If the current node’s value is $x$ or $y$, record the node’s parent node and depth. Then call the function $dfs$ on the left and right child nodes of the current node respectively, where the parent node is the current node and the depth is the current depth plus $1$. Namely $dfs(root.left, root, depth + 1)$ and $dfs(root.right, root, depth + 1)$.

When the entire binary tree is traversed, if the depths of $x$ and $y$ are the same and the parent nodes are different, $true$ will be returned, otherwise $false$ will be returned.

Time complexity $O(n)$, space complexity $O(n)$. Where $n$ is the number of nodes of the binary tree.

  • /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int x, y;
    private TreeNode p1, p2;
    private int d1, d2;

    public boolean isCousins(TreeNode root, int x, int y) {
        this.x = x;
        this.y = y;
        dfs(root, null, 0);
        return p1 != p2 && d1 == d2;
    }

    private void dfs(TreeNode root, TreeNode p, int d) {
        if (root == null) {
            return;
        }
        if (root.val == x) {
            p1 = p;
            d1 = d;
        }
        if (root.val == y) {
            p2 = p;
            d2 = d;
        }
        dfs(root.left, root, d + 1);
        dfs(root.right, root, d + 1);
    }
}


// Solution 2
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int x, y;
    private int d1, d2;
    private TreeNode p1, p2;

    public boolean isCousins(TreeNode root, int x, int y) {
        this.x = x;
        this.y = y;
        dfs(root, null, 0);
        return p1 != p2 && d1 == d2;
    }

    private void dfs(TreeNode root, TreeNode parent, int depth) {
        if (root == null) {
            return;
        }
        if (root.val == x) {
            d1 = depth;
            p1 = parent;
        } else if (root.val == y) {
            d2 = depth;
            p2 = parent;
        }
        dfs(root.left, root, depth + 1);
        dfs(root.right, root, depth + 1);
    }
}


  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isCousins(TreeNode* root, int x, int y) {
        TreeNode *p1, *p2;
        int d1, d2;
        function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* fa, int d) {
            if (!root) {
                return;
            }
            if (root->val == x) {
                p1 = fa;
                d1 = d;
            }
            if (root->val == y) {
                p2 = fa;
                d2 = d;
            }
            dfs(root->left, root, d + 1);
            dfs(root->right, root, d + 1);
        };
        dfs(root, nullptr, 0);
        return p1 != p2 && d1 == d2;
    }
};


// Solution 2
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isCousins(TreeNode* root, int x, int y) {
        int d1, d2;
        TreeNode* p1;
        TreeNode* p2;
        function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* parent, int depth) {
            if (!root) {
                return;
            }
            if (root->val == x) {
                d1 = depth;
                p1 = parent;
            } else if (root->val == y) {
                d2 = depth;
                p2 = parent;
            }
            dfs(root->left, root, depth + 1);
            dfs(root->right, root, depth + 1);
        };
        dfs(root, nullptr, 0);
        return p1 != p2 && d1 == d2;
    }
};


  • # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
        def dfs(root, fa, d):
            if root is None:
                return
            if root.val == x:
                t[0] = (fa, d)
            if root.val == y:
                t[1] = (fa, d)
            dfs(root.left, root, d + 1)
            dfs(root.right, root, d + 1)

        t = [None, None]
        dfs(root, None, 0)
        return t[0][0] != t[1][0] and t[0][1] == t[1][1]


# Solution 2
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
        def dfs(root, parent, depth):
            if root is None:
                return
            if root.val == x:
                st[0] = (parent, depth)
            elif root.val == y:
                st[1] = (parent, depth)
            dfs(root.left, root, depth + 1)
            dfs(root.right, root, depth + 1)

        st = [None, None]
        dfs(root, None, 0)
        return st[0][0] != st[1][0] and st[0][1] == st[1][1]


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isCousins(root *TreeNode, x int, y int) bool {
	var p1, p2 *TreeNode
	var d1, d2 int
	var dfs func(*TreeNode, *TreeNode, int)
	dfs = func(root *TreeNode, fa *TreeNode, d int) {
		if root == nil {
			return
		}
		if root.Val == x {
			p1, d1 = fa, d
		}
		if root.Val == y {
			p2, d2 = fa, d
		}
		dfs(root.Left, root, d+1)
		dfs(root.Right, root, d+1)
	}
	dfs(root, nil, 0)
	return p1 != p2 && d1 == d2
}


// Solution 2
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isCousins(root *TreeNode, x int, y int) bool {
	var d1, d2 int
	var p1, p2 *TreeNode
	var dfs func(root, parent *TreeNode, depth int)
	dfs = func(root, parent *TreeNode, depth int) {
		if root == nil {
			return
		}
		if root.Val == x {
			d1, p1 = depth, parent
		} else if root.Val == y {
			d2, p2 = depth, parent
		}
		dfs(root.Left, root, depth+1)
		dfs(root.Right, root, depth+1)
	}
	dfs(root, nil, 0)
	return d1 == d2 && p1 != p2
}


  • /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isCousins(root: TreeNode | null, x: number, y: number): boolean {
    let [d1, d2] = [0, 0];
    let [p1, p2] = [null, null];
    const q: [TreeNode, TreeNode][] = [[root, null]];
    for (let depth = 0; q.length > 0; ++depth) {
        const t: [TreeNode, TreeNode][] = [];
        for (const [node, parent] of q) {
            if (node.val === x) {
                [d1, p1] = [depth, parent];
            } else if (node.val === y) {
                [d2, p2] = [depth, parent];
            }
            if (node.left) {
                t.push([node.left, node]);
            }
            if (node.right) {
                t.push([node.right, node]);
            }
        }
        q.splice(0, q.length, ...t);
    }
    return d1 === d2 && p1 !== p2;
}


// Solution 2
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isCousins(root: TreeNode | null, x: number, y: number): boolean {
    let [d1, d2, p1, p2] = [0, 0, null, null];
    const dfs = (root: TreeNode | null, parent: TreeNode | null, depth: number) => {
        if (!root) {
            return;
        }
        if (root.val === x) {
            [d1, p1] = [depth, parent];
        } else if (root.val === y) {
            [d2, p2] = [depth, parent];
        }
        dfs(root.left, root, depth + 1);
        dfs(root.right, root, depth + 1);
    };
    dfs(root, null, 0);
    return d1 === d2 && p1 !== p2;
}

All Problems

All Solutions