标签：结构 ret result 输入 stat 后序 逆序输出 null ges
详细的询问：
public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null) { return null; } if (root.val > Math.max(p.val, q.val)) { return lowestCommonAncestor(root.left, p, q); } else if (root.val < Math.min(p.val, q.val)) { return lowestCommonAncestor(root.right, p, q); } else{ return root; } } } 非递归： public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { while(true){ if (root.val > Math.max(p.val, q.val)){ root = root.left; } else if (root.val < Math.min(p.val, q.val)){ root = root.right; } else{ break; } } return root; } }
2、该树有指向父节点的指针
方法1:计算两个链表长度，len = 长度之差，对齐链表，然后依次遍历比较 /** * Definition of ParentTreeNode: * * class ParentTreeNode { * public ParentTreeNode parent, left, right; * } */ public class Solution { /** * @param root: The root of the tree * @param A, B: Two node in the tree * @return: The lowest common ancestor of A and B */ public ParentTreeNode lowestCommonAncestorII(ParentTreeNode root, ParentTreeNode A, ParentTreeNode B) { int a_L = getL(A); int b_L = getL(B); int diff = a_L - b_L; ParentTreeNode l = A; ParentTreeNode s = B; if(diff < 0){ l = B; s = A; diff = b_L - a_L; } while(diff > 0) { l = l.parent; diff--; } while(l != null && s != null && l != s) { l = l.parent; s = s.parent; } return l; } public int getL(ParentTreeNode node){ int len = 0; ParentTreeNode pnode = node; while(pnode != null) { pnode = pnode.parent; len++; } return len; } } 方法2：将两个链表存入栈或数组，逆序输出比较，直到找到第一个不相等的。
3、最近公共祖先----普通树，没有指针（）
public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null) return null; if (root == p || root == q) return root; // Post order traveral TreeNode l = lowestCommonAncestor(root.left, p, q); TreeNode r = lowestCommonAncestor(root.right, p, q); if (l != null && r != null) { return root; } else { return l != null ? l : r; } }
2）可能不在树里
public class Solution { public TreeNode lowestCommonAncestor3(TreeNode root, TreeNode A, TreeNode B) { if(!covers(root,A) || !covers(root,B)){ return null; } return Helper(root, A, B); } //判断该节点node是否在当前节点root的子树里 public boolean covers(TreeNode root, TreeNode node){ if(root == null) { return false; } if(root == node){ return true;//当前节点即为node } return covers(root.left, node) || covers(root.right, node);//节点在当前节点的左右子树里 } //找公共节点 public TreeNode Helper(TreeNode root, TreeNode A, TreeNode B) { // write your code here if(root == null) { return null; } if (root == A || root == B){ return root; } boolean a_i_L = covers(root.left, A); boolean b_i_L = covers(root.left, B); //不在同一边 if(a_i_L != b_i_L){ return root; } //在通一边 TreeNode child_side = b_i_L ?root.left : root.right; return Helper(child_side, A, B); } }
方法二：构造一个结构类，两种方法。
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 { 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); } }
标签：结构 ret result 输入 stat 后序 逆序输出 null ges
原文地址：http://www.cnblogs.com/lingli-meng/p/7113369.html
