1123. The nearest common ancestor of the deepest leaf node

1123. The closest common ancestor of the deepest leaf node

Give you a root node root Two fork tree , Go back to it The nearest common ancestor of the deepest leaf node .

Think about it ：

 Leaf nodes   Is a node in a binary tree that has no child nodes 
 Of the root node of the tree   depth   by  0, If the depth of a node is  d, Then the depth of its child nodes is  d+1
 If we assume  A  It's a set of nodes  S  Of   Recent public ancestor ,S  Each node in the is represented by  A  In the subtree of the root node , And  A  The depth of reaches the maximum possible under this condition .

Example 1：

Input ：root = [3,5,1,6,2,0,8,null,null,7,4]
Output ：[2,7,4]
explain ： Our return value is 2 The node of , Mark... In yellow in the figure .
What is marked in blue in the figure is the deepest node of the tree .
Be careful , node 6、0 and 8 It's also a leaf node , But their depth is 2 , And node 7 and 4 The depth of is 3 .

Example 2：

Input ：root = [1]
Output ：[1]
explain ： The root node is the deepest node in the tree , It is its nearest common ancestor .

Example 3：

Input ：root = [0,1,3,null,2]
Output ：[2]
explain ： The deepest leaf node in the tree is 2 , The recent public ancestor is itself .

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */
 #define size 100
 int f(struct TreeNode* root){
    
     if(root){
    
         int a=f(root->left)+1;
         int b=f(root->right)+1;
         if(a>b){
    
             return a;
         }
         return b;
     }
     else{
    
         return 0;
     }
     
 }


struct TreeNode* lcaDeepestLeaves(struct TreeNode* root){
    
    struct TreeNode *queue[size],*p;
      struct TreeNode *re=NULL;
    int front,rear;
    int h=f(root);
    front=0;
    rear=0;
    queue[rear]=root;
    rear=(rear+1)%size;
     queue[rear]=NULL;
    rear=(rear+1)%size;
    printf("h %d ",h);
    int hz=1;
    if(h==1){
    
        re=root;
    }
    while(front!=rear){
    
        p=queue[front];
      
        front=(front+1)%size;
        if(p==NULL&&front!=rear){
    
            hz++;
             queue[rear]=NULL;
             rear=(rear+1)%size;
             printf("hz %d ",hz);
        }
        else{
    
            if(rear==front){
    
                break;
            }
              printf("%d |",p->val);
            if(hz==h-1){
    
                if(p->left&&p->right){
    
                    re=p;
                }
                if(p->right&&p->left==NULL){
    
                    re=p->right;
                }
                if(p->left&&p->right==NULL){
    
                    re=p->left;
                }
            }

            if(p->left){
    
                 queue[rear]=p->left;
                 rear=(rear+1)%size;
            }
            if(p->right){
    
                 queue[rear]=p->right;
                 rear=(rear+1)%size;
            }
        }
        
    }
    return re;

}

The following method should be better
1. Determine the depth of children around the current node , Equal, then the current node is the nearest common ancestor ;
2. If the depth of the left child is greater than that of the right child, the public ancestor must be on the left subtree ;
3. If the depth of the left child is less than that of the right child, the public ancestor must be on the right subtree ;
Traverse the tree and find a node with the same depth of the left and right subtrees .

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */
int getDepth(struct TreeNode* root){
    
    if(!root)return 0;
    return fmax(getDepth(root->left),getDepth(root->right))+1;
}

struct TreeNode* lcaDeepestLeaves(struct TreeNode* root){
    
    struct TreeNode* res=root,*cur=root;
    while(cur){
    
        int left=getDepth(cur->left);
        int right=getDepth(cur->right);
        if(left==right){
    
            res=cur;
            break;
        }else if(left>right){
    
            cur=cur->left;
        }else cur=cur->right;
    }
    return res;
}