LC.P979[在二叉树中分配硬币]

方法一：DFS

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/**
 * 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 {
    
    int ans;

    public int distributeCoins(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int[] dfs(TreeNode node) {
        if (node == null) return new int[]{0, 0};
        int[] left = dfs(node.left);
        int[] right = dfs(node.right);
        // 子树硬币个数
        int coins = left[0] + right[0] + node.val;
        // 子树节点数
        int nodes = left[1] + right[1] + 1;
        ans += Math.abs(coins - nodes);
        return new int[]{coins, nodes};
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：DFS(优化)

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/**
 * 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 {

    int ans;

    public int distributeCoins(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode node) {
        if (node == null) return 0;
        int d = dfs(node.left) + dfs(node.right) + node.val - 1;
        ans += Math.abs(d);
        return d;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$