Binary Tree depth

Enter a binary tree to find the depth of the tree. A tree path is formed from the root node to the leaf node. The longest path length is the depth of the tree. Input: Enter n in the first line. n indicates the number of knots. The node number ranges from 1 to n. The root node is 1. N <= 10. Next there are n rows. Each row has two integers a and B, indicating the left and right children of the I node. A is the left child, B is the right child. When a is-1, there is no left child. When B is-1, there is no right child. Output: an integer indicating the depth of the tree. Sample input: 32 3-1-1-1-1 sample output: 2 idea: There is no idea, that is, tree traversal. Code AC: [cpp] # include <stdio. h> # include <stdlib. h> typedef struct tree {int id; int vi; int lc; int rc;} tree, * p_tree; int n; p_tree t; int level; void fun (int idx, int l) {int tmp; if (t [idx]. lc =-1 & t [idx]. rc =-1) {if (l> level) {level = l ;}} else {if (t [idx]. rc <= n & t [idx]. rc> = 1) {fun (t [idx]. rc-1, l + 1);} if (t [idx]. lc <= N & t [idx]. lc> = 1) {fun (t [idx]. lc-1, l + 1) ;}} int main () {int I; while (scanf ("% d", & n )! = EOF) {www.2cto.com t = (p_tree) malloc (sizeof (tree) * n); for (I = 0; I <n; I ++) {t [I]. id = I + 1; scanf ("% d", & t [I]. lc, & t [I]. rc);} level =-1; fun (0, 1); printf ("% d \ n", level);} return 0 ;}