2641. 二叉树的堂兄弟节点 II

2641. 二叉树的堂兄弟节点 II

• 难度: medium
• 原始链接: https://leetcode.cn/problems/cousins-in-binary-tree-ii
• 标签: 广度优先遍历

解法一: 广度优先遍历

go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func replaceValueInTree(root *TreeNode) *TreeNode {
  root.Val = 0
  queue := []*TreeNode{root}
  for len(queue) > 0 {
    sonSum := 0
    for _, node := range queue {
      if node.Left != nil {
        sonSum += node.Left.Val
      }
      if node.Right != nil {
        sonSum += node.Right.Val
      }
    }
    size := len(queue)
    for _, node := range queue[:size] {
      curSonSum := 0
      if node.Left != nil {
        curSonSum += node.Left.Val
      }
      if node.Right != nil {
        curSonSum += node.Right.Val
      }
      curSonVal := sonSum - curSonSum
      if node.Left != nil {
        node.Left.Val = curSonVal
        queue = append(queue, node.Left)
      }
      if node.Right != nil {
        node.Right.Val = curSonVal
        queue = append(queue, node.Right)
      }
    }
    queue = queue[size:]
  }
  return root
}

java

/**
 * 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 {
    public TreeNode replaceValueInTree(TreeNode root) {
        root.val = 0;
        Queue<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int sonSum = 0;
            for (TreeNode node : queue) {
                if (node.left != null) {
                    sonSum += node.left.val;
                }
                if (node.right != null) {
                    sonSum += node.right.val;
                }
            }
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int curSonSum = 0;
                TreeNode node = queue.poll();
                if (node.left != null) {
                    curSonSum += node.left.val;
                }
                if (node.right != null) {
                    curSonSum += node.right.val;
                }
                int curSonVal = sonSum - curSonSum;
                if (node.left != null) {
                    node.left.val = curSonVal;
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    node.right.val = curSonVal;
                    queue.offer(node.right);
                }
            }
        }
        return root;
    }
}