563. 二叉树的坡度

563. 二叉树的坡度

原题

/**
 * 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 res = 0;
    public int findTilt(TreeNode root) {
        if(root==null){
            return 0;
        }
        dfs(root);
        return res;
    }
    public int dfs(TreeNode root){
        if(root==null){
            return 0;
        }
        //左子树遍历递归
        int left = dfs(root.left);
        //右子树遍历递归
        int right = dfs(root.right);
        //求累计坡度
        res += Math.abs(left-right);
        //计算当前结点和其子树的总值
        return root.val+left+right;
    }
}

你可能感兴趣的:(LeetCode刷题,数据结构,深度优先)