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2641. Cousins in Binary Tree II

Description

Given the root of a binary tree, replace the value of each node in the tree with the sum of all its cousins' values.

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

Return the root of the modified tree.

Note that the depth of a node is the number of edges in the path from the root node to it.

 

Example 1:

Input: root = [5,4,9,1,10,null,7]
Output: [0,0,0,7,7,null,11]
Explanation: The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 5 does not have any cousins so its sum is 0.
- Node with value 4 does not have any cousins so its sum is 0.
- Node with value 9 does not have any cousins so its sum is 0.
- Node with value 1 has a cousin with value 7 so its sum is 7.
- Node with value 10 has a cousin with value 7 so its sum is 7.
- Node with value 7 has cousins with values 1 and 10 so its sum is 11.

Example 2:

Input: root = [3,1,2]
Output: [0,0,0]
Explanation: The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 3 does not have any cousins so its sum is 0.
- Node with value 1 does not have any cousins so its sum is 0.
- Node with value 2 does not have any cousins so its sum is 0.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 105].
  • 1 <= Node.val <= 104

Solutions

  • /**
 * 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 List<Integer> s = new ArrayList<>();

    public TreeNode replaceValueInTree(TreeNode root) {
        dfs1(root, 0);
        root.val = 0;
        dfs2(root, 1);
        return root;
    }

    private void dfs1(TreeNode root, int d) {
        if (root == null) {
            return;
        }
        if (s.size() <= d) {
            s.add(0);
        }
        s.set(d, s.get(d) + root.val);
        dfs1(root.left, d + 1);
        dfs1(root.right, d + 1);
    }

    private void dfs2(TreeNode root, int d) {
        if (root == null) {
            return;
        }
        int l = root.left == null ? 0 : root.left.val;
        int r = root.right == null ? 0 : root.right.val;
        if (root.left != null) {
            root.left.val = s.get(d) - l - r;
        }
        if (root.right != null) {
            root.right.val = s.get(d) - l - r;
        }
        dfs2(root.left, d + 1);
        dfs2(root.right, d + 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:
    TreeNode* replaceValueInTree(TreeNode* root) {
        vector<int> s;
        function<void(TreeNode*, int)> dfs1 = [&](TreeNode* root, int d) {
            if (!root) {
                return;
            }
            if (s.size() <= d) {
                s.push_back(0);
            }
            s[d] += root->val;
            dfs1(root->left, d + 1);
            dfs1(root->right, d + 1);
        };
        function<void(TreeNode*, int)> dfs2 = [&](TreeNode* root, int d) {
            if (!root) {
                return;
            }
            int l = root->left ? root->left->val : 0;
            int r = root->right ? root->right->val : 0;
            if (root->left) {
                root->left->val = s[d] - l - r;
            }
            if (root->right) {
                root->right->val = s[d] - l - r;
            }
            dfs2(root->left, d + 1);
            dfs2(root->right, d + 1);
        };
        dfs1(root, 0);
        root->val = 0;
        dfs2(root, 1);
        return root;
    }
};

  • # 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 replaceValueInTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs1(root, d):
            if root is None:
                return
            if len(s) <= d:
                s.append(0)
            s[d] += root.val
            dfs1(root.left, d + 1)
            dfs1(root.right, d + 1)

        def dfs2(root, d):
            if root is None:
                return
            t = (root.left.val if root.left else 0) + (
                root.right.val if root.right else 0
            )
            if root.left:
                root.left.val = s[d] - t
            if root.right:
                root.right.val = s[d] - t
            dfs2(root.left, d + 1)
            dfs2(root.right, d + 1)

        s = []
        dfs1(root, 0)
        root.val = 0
        dfs2(root, 1)
        return root


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func replaceValueInTree(root *TreeNode) *TreeNode {
	s := []int{}
	var dfs1 func(*TreeNode, int)
	dfs1 = func(root *TreeNode, d int) {
		if root == nil {
			return
		}
		if len(s) <= d {
			s = append(s, 0)
		}
		s[d] += root.Val
		dfs1(root.Left, d+1)
		dfs1(root.Right, d+1)
	}
	var dfs2 func(*TreeNode, int)
	dfs2 = func(root *TreeNode, d int) {
		if root == nil {
			return
		}
		l, r := 0, 0
		if root.Left != nil {
			l = root.Left.Val
		}
		if root.Right != nil {
			r = root.Right.Val
		}
		if root.Left != nil {
			root.Left.Val = s[d] - l - r
		}
		if root.Right != nil {
			root.Right.Val = s[d] - l - r
		}
		dfs2(root.Left, d+1)
		dfs2(root.Right, d+1)
	}
	dfs1(root, 0)
	root.Val = 0
	dfs2(root, 1)
	return root
}

  • /**
 * 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 replaceValueInTree(root: TreeNode | null): TreeNode | null {
    const s: number[] = [];
    const dfs1 = (root: TreeNode | null, d: number): void => {
        if (!root) {
            return;
        }
        if (s.length <= d) {
            s.push(0);
        }
        s[d] += root.val;
        dfs1(root.left, d + 1);
        dfs1(root.right, d + 1);
    };
    const dfs2 = (root: TreeNode | null, d: number): void => {
        if (!root) {
            return;
        }
        const t = (root.left?.val ?? 0) + (root.right?.val ?? 0);
        if (root.left) {
            root.left.val = s[d] - t;
        }
        if (root.right) {
            root.right.val = s[d] - t;
        }
        dfs2(root.left, d + 1);
        dfs2(root.right, d + 1);
    };
    dfs1(root, 0);
    root.val = 0;
    dfs2(root, 1);
    return root;
}

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