平衡二叉树性质：求解《面试题 04.06. 后继者》

var inorderSuccessor = function(root, p) {
  let r = null
  while (root) {
    if (root.val > p.val) {
      r = root
      root = root.left
    } else {
      root = root.right
    }
  }
  return r
};

func inorderSuccessor(root *TreeNode, p *TreeNode) *TreeNode {
  var r *TreeNode
  for root != nil {
    if root.Val > p.Val {
      r = root
      root = root.Left
    } else {
      root = root.Right
    }
  }
  return r
}

class Solution {
  public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
    TreeNode r = null;
    while (root != null) {
      if (root.val > p.val) {
        r = root;
        root = root.left;
      } else {
        root = root.right;
      }
    }
    return r;
  }
}

class Solution:
  def inorderSuccessor(self, root: TreeNode, p: TreeNode) -> TreeNode:
    r = None
    while root:
      if root.val > p.val:
        r = root
        root = root.left
      else:
        root = root.right 
    return r

def inorder_successor(root, p)
  r = nil
  while root do
    if root.val > p.val 
      r = root
      root = root.left
    else
      root = root.right
    end
  end
  return r
end

struct TreeNode* inorderSuccessor(struct TreeNode* root, struct TreeNode* p){
  struct TreeNode *r = NULL;
  while (root != NULL) {
    if (root->val > p->val) {
      r = root;
      root = root->left;
    } else {
      root = root->right;
    }
  }
  return r;
}

class Solution {
public:
  TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
    TreeNode* r = nullptr;
    while (root != nullptr) {
      if (root->val > p->val) {
        r = root;
        root = root->left;
      } else {
        root = root->right;
      }
    }
    return r;
  }
};

public class Solution {
  public TreeNode InorderSuccessor(TreeNode root, TreeNode p) {
    TreeNode r = null;
    while (root != null) {
      if (root.val > p.val) {
        r = root;
        root = root.left;
      } else {
        root = root.right;
      }
    }
    return r;
  }
}