标签:HSM color rom The nan treenode add lin root
902. Kth Smallest Element in a BST
https://www.lintcode.com/problem/kth-smallest-element-in-a-bst/description?_from=ladder&&fromId=1
public class Solution { /** * @param root: the given BST * @param k: the given k * @return: the kth smallest element in BST */ public int kthSmallest(TreeNode root, int k) { // write your code here Map<TreeNode, Integer> numOfChildren = new HashMap<>(); countNodes(root, numOfChildren); return quickSelectOnTree(numOfChildren, k, root); } public int countNodes(TreeNode root, Map<TreeNode, Integer> numOfChildren) { if(root == null) { return 0; } int left = countNodes(root.left, numOfChildren); int right = countNodes(root.right, numOfChildren); numOfChildren.put(root, left + right + 1); return left + right + 1; } public int quickSelectOnTree(Map<TreeNode, Integer> numOfChildren, int k, TreeNode root) { if(root == null) { return -1; } int left = root.left == null ? 0 : numOfChildren.get(root.left); if(left >= k) { return quickSelectOnTree(numOfChildren, k, root.left); } if(left + 1 == k) { return root.val; } return quickSelectOnTree(numOfChildren, k - left - 1, root.right); } }
578. Lowest Common Ancestor III
https://www.lintcode.com/problem/lowest-common-ancestor-iii/description?_from=ladder&&fromId=1
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ class ResultType { public boolean a_exist, b_exist; public TreeNode node; ResultType(boolean a, boolean b, TreeNode n) { a_exist = a; b_exist = b; node = n; } } public class Solution { /* * @param root: The root of the binary tree. * @param A: A TreeNode * @param B: A TreeNode * @return: Return the LCA of the two nodes. */ public TreeNode lowestCommonAncestor3(TreeNode root, TreeNode A, TreeNode B) { // write your code here ResultType rt = helper(root, A, B); if(rt.a_exist && rt.b_exist) { return rt.node; } else { return null; } } public ResultType helper(TreeNode root, TreeNode A, TreeNode B) { if(root == null) { return new ResultType(false, false, null); } ResultType left_rt = helper(root.left, A, B); ResultType right_rt = helper(root.right, A, B); boolean a_exist = left_rt.a_exist || right_rt.a_exist || root == A; boolean b_exist = left_rt.b_exist || right_rt.b_exist || root == B; if(root == A || root == B) { return new ResultType(a_exist, b_exist, root); } if(left_rt.node != null && right_rt.node != null) { return new ResultType(a_exist, b_exist, root); } if(left_rt.node != null) { return new ResultType(a_exist, b_exist, left_rt.node); } if(right_rt.node != null) { return new ResultType(a_exist, b_exist, right_rt.node); } return new ResultType(a_exist, b_exist, null); } }
Binary Tree - Divide Conquer & Traverse
标签:HSM color rom The nan treenode add lin root
原文地址:https://www.cnblogs.com/jenna/p/10921321.html