Minimum Depth of Binary Tree 二叉树最小深度

//运用递归的思想，当节点为空时间，即1 2 V 3 V V V 。当左子树为空时，及左子树在具象图上面根本不存在，此时root的深度等于右子树的深度+1；反之亦然

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public int minDepth(TreeNode root) {          return minRec(root);      }            private int minRec(TreeNode root) {          if(root==null) return 0;                    int l = minRec(root.left);          int r = minRec(root.right);                    if(l==0) return r+1;          if(r==0) return l+1;                    return Math.min(l, r) + 1;                }  }  

//也可以从另一方面想，就是当节点为空时间，判断右子树是否存在，当存在时，此节点深度为正无穷（INT_MAX）；若不存在，则此节点深度为0

public class MinimumDepthofBinaryTree {      public int minDepth(TreeNode root) {          if (root == null)              return 0;          if (root.left == null && root.right == null)             return 1;         else {             int leftDepth = root.left != null ? minDepth(root.left)                     : Integer.MAX_VALUE;             int rightDepth = root.right != null ? minDepth(root.right)                     : Integer.MAX_VALUE;             return Math.min(leftDepth,rightDepth) + 1;         }     }}

int minDepth(TreeNode *root) {          // Start typing your C/C++ solution below          // DO NOT write int main() function          if(root == NULL)          {              return 0;          }          else          {              minD(root);          }      }            int minD(TreeNode *root) {          if(root->left == NULL && root->right == NULL)          {              return 1;          }          else          {              if(root->left == NULL)              {                  return minD(root->right) + 1;              }              else if(root->right == NULL)              {                  return minD(root->left) + 1;              }              else              {                  int left = minD(root->left);                  int right = minD(root->right);                  return (left<right?left:right) + 1;              }          }      }  };