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Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).” _______6______ / ___2__ ___8__ / \ / 0 _4 7 9 / 3 5 For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
思路，先判断入口是否有非法输入。
1 如果某一个root==p || root == q，那么LCA肯定是root（因为是top down，LCA肯定在root所囊括的树上，而root又是p q其中一个节点了，那么另外一个节点肯定在root之下，那么root就是LCA），那么返回root
2 如果root<min(p, q)，那么LCA肯定在右子树上，那么递归
3 如果max(p, q)<root，那么LCA肯定在左子树上，那么递归
4 如果p<root<q，那么root肯定为LCA
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
/*
if(p == root || q == root || root == null){
return root;
}
TreeNode left=lowestCommonAncestor(root.left, p, q);
TreeNode right=lowestCommonAncestor(root.right, p, q);
//在两边
if(left != null && right != null ) return root;
//在一边
if(left != null){
}
*/
if(root==null || root == p || root == q){
return root;
}
if(root.val < Math.min(p.val, q.val)){
return lowestCommonAncestor(root.right, p, q);
}
else if(root.val > Math.max(p.val, q.val)){
return lowestCommonAncestor(root.left, p, q);
}
else {
return root;
}
}
}
LeetCode-Lowest Common Ancestor of a Binary Search Tree
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原文地址：http://www.cnblogs.com/incrediblechangshuo/p/5548290.html
