1372. Longest ZigZag Path in a Binary Tree

减掉一个分支，否则会超时的。已经走过的路就不再走了。

/**
 * 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:
    
    int func(TreeNode* root, char direction){
        if (root==NULL){
            return 0;
        }
        
        if (direction=='L'){
            return func(root->left, 'R')+1;
        }
        else{
            return func(root->right, 'L')+1;
        }
    }
    
    
    int longestZigZag(TreeNode* root, bool dicleft=true, bool dicright=true) {
        
        if (root==NULL){
            return 0;
        }
        
        if (root->left==NULL&&root->right==NULL){
            return 0;
        }
        
        int max1 = 0;
        int max2 = 0;
        if (dicleft){
            max1 = func(root->left, 'R');
        }
        if (dicright){
            max2 = func(root->right, 'L');
        }
        int max = max1 > max2? max1: max2;
        
        int mmax1 = longestZigZag(root->left, true, false);
        int mmax2 = longestZigZag(root->right, false, true);
        max = max > mmax1? max: mmax1;
        max = max > mmax2? max: mmax2;
        
        return max;
        
    }
};

标签：Binary,right,return,int,max,Tree,ZigZag,NULL,root
来源： https://blog.csdn.net/zeroQiaoba/article/details/104758387