[leetcode] Lowest Common Ancestor of a Binary Tree

最新推荐文章于 2024-10-16 01:06:12 发布

Cabinathor

最新推荐文章于 2024-10-16 01:06:12 发布

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本文介绍了一种寻找二叉树中两个指定节点的最低公共祖先(LCA)的方法。通过递归路径查找的方式，记录从根节点到目标节点的路径，并对比两条路径找到最后一个相同的节点即为所求。

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From : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

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.

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(E) Lowest Common Ancestor of a Binary Search Tree

Solution :

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool getPath(TreeNode* root, TreeNode* p, vector<TreeNode *> &path) {
        if(!root) return false;
        path.push_back(root);
        if(root == p) return true;
        if(getPath(root->left, p, path) || getPath(root->right, p, path)) return true;
        path.pop_back();
        return false;
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        vector<TreeNode *> pPath, qPath;
        getPath(root, p ,pPath);
        getPath(root, q, qPath);
        TreeNode *lca = NULL;
        for(int i=0, lp=pPath.size(), lq=qPath.size(); i<lp && i<lq; i++) {
            if(pPath[i] == qPath[i]) {
                lca = pPath[i];
            } else {
                break;
            }
        }
        return lca;
    }
};

Solution :

其实对p和q的getPath可以一起进行。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void getPath(TreeNode* root, TreeNode* p, TreeNode* q, vector<TreeNode *> &path, vector<TreeNode *> &pPath, vector<TreeNode *> &qPath) {
        if(!root) return;
        path.push_back(root);
        if(root == p) pPath = path;
        if(root == q) qPath = path;
        if(!pPath.empty() && !qPath.empty()) return;
        getPath(root->left, p, q, path, pPath, qPath);
        getPath(root->right, p, q, path, pPath, qPath);
        path.pop_back();
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        vector<TreeNode *> pPath, qPath, path;
        getPath(root, p, q, path, pPath, qPath);
        TreeNode *lca = NULL;
        for(int i=0, lp=pPath.size(), lq=qPath.size(); i<lp && i<lq; i++) {
            if(pPath[i] == qPath[i]) {
                lca = pPath[i];
            } else {
                break;
            }
        }
        return lca;
    }
};