563. Binary Tree Tilt

最新推荐文章于 2019-03-02 23:18:29 发布

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最新推荐文章于 2019-03-02 23:18:29 发布

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分类专栏： LeetCode刷刷刷

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本文介绍了一种计算二叉树每个节点的倾斜值并求得整棵树总倾斜的方法。倾斜值定义为节点左子树与右子树节点值之和的绝对差。通过递归遍历整棵树，计算各节点倾斜值并累加得到最终结果。

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Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes' tilt.

Example:

Input: 
         1
       /   \
      2     3
Output: 1
Explanation: 
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

The sum of node values in any subtree won't exceed the range of 32-bit integer.
All the tilt values won't exceed the range of 32-bit integer.

題意:

返回整個樹的「傾斜」，所謂傾斜就是將左子樹總和與右子樹總和的差，例如：

1 節點1的傾斜: |14-9|=5
/ \ 節點2的傾斜: |7-5|=2
2 3 節點3的傾斜: |6|=6
/ \ \ 節點5的傾斜: |0|=0
7 5 6 節點6的傾斜: |0|=0

節點7的傾斜: |0|=0

整棵樹的傾斜:5+2+6+0+0+0=13

題解:

歷遍整棵樹，在歷遍每個節點時，都計算傾斜，並存儲在ArrayList中，然後當前子樹的總值(計算左右子樹加上自己的值)，在返回。

歷遍完後，再把所有傾斜做加總，再返回答案。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class BinaryTreeTilt {
    public int findTilt(TreeNode root) {
        if(root == null)
            return 0;
        
        List<Integer> tilts = new ArrayList<>();
        helper(root, tilts);
        
        int result = 0;
        for(int i = 0; i < tilts.size(); i ++) {
            result += tilts.get(i);
        }
        
        return result;
    }
    
    int helper(TreeNode root, List<Integer> tilts) {
        if(root == null)
            return 0;
        
        if(root.left == null && root.right == null)
            return root.val;
        
        int left = 0;
        if(root.left != null)
            left += helper(root.left, tilts);
        
        int right = 0;
        if(root.right != null)
            right += helper(root.right, tilts);
        
        int tilt = Math.abs(left - right);
        tilts.add(tilt);
        
        int curSum = left + right + root.val;
        
        return curSum;
    }
}