leetcode 1372

问题

Given a binary tree root, a ZigZag path for a binary tree is defined as follow:

  • Choose any node in the binary tree and a direction (right or left).
  • If the current direction is right then move to the right child of the current node otherwise move to the left child.
  • Change the direction from right to left or right to left.
  • Repeat the second and third step until you can’t move in the tree.

Zigzag length is defined as the number of nodes visited - 1. (A single node has a length of 0).

Return the longest ZigZag path contained in that tree.

Example 1:

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Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1,null,1]
Output: 3
Explanation: Longest ZigZag path in blue nodes (right -> left -> right).

Example 2:

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Input: root = [1,1,1,null,1,null,null,1,1,null,1]
Output: 4
Explanation: Longest ZigZag path in blue nodes (left -> right -> left -> right).

Example 3:

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Input: root = [1]
Output: 0

Constraints:

  • Each tree has at most 50000 nodes..
  • Each node’s value is between [1, 100].

分析

注意dp可以有多个返回值，要对每个节点进行处理，回溯时在进行总和。

代码

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    // [0]: as left  in
    // [1]: as right in
    vector<int> dp(TreeNode *root) {
        if (!root) return {0, 0, 0};
        vector<int> l = dp(root->left);
        vector<int> r = dp(root->right);
        vector<int> ret(3, 0);
        ret[0] = root->right? r[1]+1: 0;
        ret[1] = root->left?  l[0]+1: 0;
        ret[2] = max(max(ret[0], ret[1]), max(l[2], r[2]));
        return ret;
    }
    int longestZigZag(TreeNode* root) {
        return dp(root)[2];
    }
};