236. Lowest Common Ancestor of a Binary Tree

题目：

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

For example, the lowest common ancestor (LCA) of nodes 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.

链接： http://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

题解：

普通二叉树求公共祖先。看过<剑指Offer>以后知道这道题应该形成一系列问题。比如是不是二叉树，是不是BST。假如是BST的话我们可以用上题的方法，二分搜索。有没有指向父节点的link，假如有指向父节点的link我们就可以用intersection of two lists的方法找到两个linked list相交的地方。 对这道题目，我们使用后续遍历来做：

1. 定义两个辅助节点，使用后续遍历来遍历整个树
2. 当root的值等于p或者q时，找到一个符合条件的节点，返回这个root
3. 先遍历左子树
4. 再遍历右子树
5. 当left，right均找到时返回此root
6. 只找到left时返回left
7. 只找到right时返回right
8. 否则返回null

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {
    public static 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);      // Post order traveral
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        
        if (left != null && right != null)          // p and q in two subtrees
            return root;
        else
            return left != null ? left : right; 
    }
}

 

二刷:

Java:

/**
 * 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 || root == p || root == q) return root;
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left != null && right != null) return root;
        else return left != null ? left : right;
    }
}

 

三刷

还是跟以前一样的方法，利用递归，先遍历两个子树，来查找是否其中含有目标节点p或者q。假如两节点分别位于root的左右两侧，则root为LCA. 否则，left和right哪个非空，则哪一个为LCA, 这一侧含有p和q两个目标节点。

Java:

/**
 * 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);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left != null && right != null) return root;
        else return (left != null) ? left : right;
    }
}

 

 

 

Reference: