701. 二叉搜索树中的插入操作

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递归

  1. /**
  2. * Definition for a binary tree node.
  3. * public class TreeNode {
  4. * int val;
  5. * TreeNode left;
  6. * TreeNode right;
  7. * TreeNode() {}
  8. * TreeNode(int val) { this.val = val; }
  9. * TreeNode(int val, TreeNode left, TreeNode right) {
  10. * this.val = val;
  11. * this.left = left;
  12. * this.right = right;
  13. * }
  14. * }
  15. */
  16. class Solution {
  17. public TreeNode insertIntoBST(TreeNode root, int val) {
  18. //递归有返回值
  19. if(root == null ) {
  20. TreeNode node = new TreeNode(val);
  21. return node;
  22. }
  23. if(root.val > val ) {
  24. root.left = insertIntoBST(root.left,val);
  25. }
  26. if(root.val < val ) {
  27. root.right = insertIntoBST(root.right,val);
  28. }
  29. return root;
  30. }
  31. }

递归无返回值

  1. /**
  2. * Definition for a binary tree node.
  3. * public class TreeNode {
  4. * int val;
  5. * TreeNode left;
  6. * TreeNode right;
  7. * TreeNode() {}
  8. * TreeNode(int val) { this.val = val; }
  9. * TreeNode(int val, TreeNode left, TreeNode right) {
  10. * this.val = val;
  11. * this.left = left;
  12. * this.right = right;
  13. * }
  14. * }
  15. */
  16. class Solution {
  17. TreeNode parent;
  18. public TreeNode insertIntoBST(TreeNode root, int val) {
  19. parent = new TreeNode(0);
  20. if(root == null ) {
  21. root = new TreeNode(val);
  22. }
  23. traversal(root, val );
  24. return root;
  25. }
  26. public void traversal(TreeNode root, int val ) {
  27. if(root == null ) {
  28. TreeNode node = new TreeNode(val);
  29. if(val > parent.val ) parent.right = node;
  30. else parent.left = node;
  31. return;
  32. }
  33. parent = root;
  34. if(root.val > val ) traversal(root.left,val);
  35. if(root.val < val ) traversal(root.right,val);
  36. return ;
  37. }
  38. }

迭代

  1. /**
  2. * Definition for a binary tree node.
  3. * public class TreeNode {
  4. * int val;
  5. * TreeNode left;
  6. * TreeNode right;
  7. * TreeNode() {}
  8. * TreeNode(int val) { this.val = val; }
  9. * TreeNode(int val, TreeNode left, TreeNode right) {
  10. * this.val = val;
  11. * this.left = left;
  12. * this.right = right;
  13. * }
  14. * }
  15. */
  16. class Solution {
  17. public TreeNode insertIntoBST(TreeNode root, int val) {
  18. //迭代
  19. if(root == null ) {
  20. TreeNode node = new TreeNode(val);
  21. return node;
  22. }
  23. TreeNode cur = root;
  24. TreeNode parent = root;
  25. while( cur != null ) {
  26. parent = cur;
  27. if(cur.val > val) {
  28. cur = cur.left;
  29. } else {
  30. cur = cur.right;
  31. }
  32. }
  33. TreeNode node = new TreeNode(val);
  34. if(parent.val > val ) {
  35. parent.left = node;
  36. } else {
  37. parent.right = node;
  38. }
  39. return root;
  40. }
  41. }