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一、Description
题目描述:给定一个二叉树和两个结点p和q,找出p和q的最近公共祖先。
二、Analyzation
通过遍历分别找到从根节点到p和q的路径,放入一个栈中。如果两个栈的大小相同,则同时出栈直到两个栈出栈的结点值相同,则为最近公共祖先,如果两个栈的大小不同(p和q不在同一层),则大的那个栈出栈直到和另一个栈的大小相同,然后一起出栈直到结点值相同。
三、Accepted code
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || p == null || q == null) {
return null;
}
Stack<TreeNode> stack1 = new Stack<>();
Stack<TreeNode> stack2 = new Stack<>();
help(root, p, stack1);
help(root, q, stack2);
if (stack1.size() > stack2.size()) {
while (stack1.size() != stack2.size()) {
stack1.pop();
}
} else if (stack1.size() < stack2.size()) {
while (stack1.size() != stack2.size()) {
stack2.pop();
}
}
while (!stack1.isEmpty() && !stack2.isEmpty()) {
if (stack1.peek().val == stack2.peek().val) {
return stack1.peek();
}
stack1.pop();
stack2.pop();
}
return null;
}
public boolean help(TreeNode root, TreeNode node, Stack<TreeNode> stack) {
stack.add(root);
if (root.val == node.val) {
return true;
}
boolean find = false;
if (root.left != null) {
find = help(root.left, node, stack);
}
if (root.right != null && !find) {
find = help(root.right, node, stack);
}
if (!find) {
stack.pop();
}
return find;
}
}