Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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).”
_______3______
/ ___5__ ___1__
/ \ / 6 _2 0 8
/ 7 4
5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition./**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
TreeNode node = null;
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
findNode(root, p, q);
return node;
}
public int findNode(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return 0;
}
int re = 0;
if (root == p) {
re = 1 + findNode(root.left, p, q) + findNode(root.right, p, q);
} else if (root == q) {
re = 2 + findNode(root.left, p, q) + findNode(root.right, p, q);
} else {
int left = findNode(root.left, p, q);
if (left == 3) {
return 3;
}
int right = findNode(root.right, p, q);
if (right == 3) {
return 3;
}
re = left + right;
}
if (re == 3) {
node = root;
}
return re;
}
}
recursively root,检查root是否是pq，root的左子树是否有pq，root的右子树是否有pq。 18%
**这种办法是基于p、q一定在ｒｏｏｔ中
/**
* 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(root==null) return null;
if(root==p || root==q) return root;
TreeNode left=lowestCommonAncestor(root.left, p, q), right = lowestCommonAncestor(root.right, p, q);
if(left!=null && right!=null) return root;
if(right==null) return left;
if(left==null) return right;
return null;
}
}
236. Lowest Common Ancestor of a Binary Tree
原文：http://www.cnblogs.com/yuchenkit/p/7192628.html