设计并实现一个算法，找出二叉树中某两个节点的第一个共同祖先。不得将其他的节点存储在另外的数据结构中。注意：这不一定是二叉搜索树。
例如，给定如下二叉树: root = [3,5,1,6,2,0,8,null,null,7,4]
3
/ \
5 1
/ \ / \
6 2 0 8
/ \
7 4示例 1:
输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
输出: 3
解释: 节点 5 和节点 1 的最近公共祖先是节点 3。示例 2:
输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
输出: 5
解释: 节点 5 和节点 4 的最近公共祖先是节点 5。因为根据定义最近公共祖先节点可以为节点本身。说明:
所有节点的值都是唯一的。
p、q 为不同节点且均存在于给定的二叉树中。
//leetcode submit region begin(Prohibit modification and deletion)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || p == null || q == null) {
return null;
}
Map<TreeNode, TreeNode> parentMap = new HashMap<>();
createParentMap(parentMap, root);
Stack<TreeNode> pStack = new Stack<>();
Stack<TreeNode> qStack = new Stack<>();
TreeNode tempNode = p;
while (parentMap.get(tempNode) != null) {
pStack.push(tempNode);
tempNode = parentMap.get(tempNode);
}
pStack.push(root);
tempNode = q;
while (parentMap.get(tempNode) != null) {
qStack.push(tempNode);
tempNode = parentMap.get(tempNode);
}
qStack.push(root);
TreeNode result = null;
while (!pStack.isEmpty() && !qStack.isEmpty() && pStack.peek() == qStack.peek()) {
result = pStack.peek();
pStack.pop();
qStack.pop();
}
return result;
}
private void createParentMap(Map<TreeNode, TreeNode> parentMap, TreeNode node) {
if (node == null) {
return;
}
if (node.left != null) {
parentMap.put(node.left, node);
}
if (node.right != null) {
parentMap.put(node.right, node);
}
createParentMap(parentMap, node.left);
createParentMap(parentMap, node.right);
}
}
//leetcode submit region end(Prohibit modification and deletion)
